Find all parameters $a,b,c,d in mathbb R$ for which the function $f: mathbb R rightarrow mathbb R$ a) has a...












0














My function $f: mathbb R rightarrow mathbb R$ it is given a pattern:
$$f(x)=begin{cases} frac{2^{7^{x}-1}-a}{ln(1-x)},& x<0\ b,& x=0 \ frac{sinleft(c sqrt{x^{2}+d^{2}x}right) }{x},& x>0 end{cases}$$ for $$a,b,c,d in mathbb R$$
I have a problem with this task because function patterns for $x</>0$ have they reach zero in the denominator for $x=0.$



That is why I think that I should cut a denominator with a numerator but I cannot do it because I have logarithm and sine so I need other way to this task.



I thought about doing some substitution: for example $y=frac{1}{x}$, then for $x rightarrow 0^mp$ I have $y rightarrow pm infty$ and but I still can not solve this.



Can I count on any tips?






real-analysis







share|cite|improve this question
























  • so the goal is to find the parameters such that the function defined for x negative, x positive and x = 0 is continuous ?
    – Marine Galantin
    2 days ago










  • @MP3129 Are you familiar with Taylor's Theorem?
    – Cameron Buie
    2 days ago










  • @MarineGalantin in point (a) I need to find the parameters $ a, b, c, d in mathbb R $ for which the function has a limit in point 0 and in point (b) which is a separate case I have to indicate parameters for which the function is continuous in the whole field
    – MP3129
    2 days ago












  • @CameronBuie I have not a Taylor's Theorem at my university and I mustn't use facts which I did not have a lecture
    – MP3129
    2 days ago






  • 1




    @MP3129: Okay. What about L'Hôpital's Rule?
    – Cameron Buie
    2 days ago
















0














My function $f: mathbb R rightarrow mathbb R$ it is given a pattern:
$$f(x)=begin{cases} frac{2^{7^{x}-1}-a}{ln(1-x)},& x<0\ b,& x=0 \ frac{sinleft(c sqrt{x^{2}+d^{2}x}right) }{x},& x>0 end{cases}$$ for $$a,b,c,d in mathbb R$$
I have a problem with this task because function patterns for $x</>0$ have they reach zero in the denominator for $x=0.$



That is why I think that I should cut a denominator with a numerator but I cannot do it because I have logarithm and sine so I need other way to this task.



I thought about doing some substitution: for example $y=frac{1}{x}$, then for $x rightarrow 0^mp$ I have $y rightarrow pm infty$ and but I still can not solve this.



Can I count on any tips?






real-analysis







share|cite|improve this question
























  • so the goal is to find the parameters such that the function defined for x negative, x positive and x = 0 is continuous ?
    – Marine Galantin
    2 days ago










  • @MP3129 Are you familiar with Taylor's Theorem?
    – Cameron Buie
    2 days ago










  • @MarineGalantin in point (a) I need to find the parameters $ a, b, c, d in mathbb R $ for which the function has a limit in point 0 and in point (b) which is a separate case I have to indicate parameters for which the function is continuous in the whole field
    – MP3129
    2 days ago












  • @CameronBuie I have not a Taylor's Theorem at my university and I mustn't use facts which I did not have a lecture
    – MP3129
    2 days ago






  • 1




    @MP3129: Okay. What about L'Hôpital's Rule?
    – Cameron Buie
    2 days ago














0












0








0







My function $f: mathbb R rightarrow mathbb R$ it is given a pattern:
$$f(x)=begin{cases} frac{2^{7^{x}-1}-a}{ln(1-x)},& x<0\ b,& x=0 \ frac{sinleft(c sqrt{x^{2}+d^{2}x}right) }{x},& x>0 end{cases}$$ for $$a,b,c,d in mathbb R$$
I have a problem with this task because function patterns for $x</>0$ have they reach zero in the denominator for $x=0.$



That is why I think that I should cut a denominator with a numerator but I cannot do it because I have logarithm and sine so I need other way to this task.



I thought about doing some substitution: for example $y=frac{1}{x}$, then for $x rightarrow 0^mp$ I have $y rightarrow pm infty$ and but I still can not solve this.



Can I count on any tips?






real-analysis







share|cite|improve this question















My function $f: mathbb R rightarrow mathbb R$ it is given a pattern:
$$f(x)=begin{cases} frac{2^{7^{x}-1}-a}{ln(1-x)},& x<0\ b,& x=0 \ frac{sinleft(c sqrt{x^{2}+d^{2}x}right) }{x},& x>0 end{cases}$$ for $$a,b,c,d in mathbb R$$
I have a problem with this task because function patterns for $x</>0$ have they reach zero in the denominator for $x=0.$



That is why I think that I should cut a denominator with a numerator but I cannot do it because I have logarithm and sine so I need other way to this task.



I thought about doing some substitution: for example $y=frac{1}{x}$, then for $x rightarrow 0^mp$ I have $y rightarrow pm infty$ and but I still can not solve this.



Can I count on any tips?






real-analysis




real-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Cameron Buie

85.1k771155




85.1k771155










asked 2 days ago









MP3129MP3129

876




876












  • so the goal is to find the parameters such that the function defined for x negative, x positive and x = 0 is continuous ?
    – Marine Galantin
    2 days ago










  • @MP3129 Are you familiar with Taylor's Theorem?
    – Cameron Buie
    2 days ago










  • @MarineGalantin in point (a) I need to find the parameters $ a, b, c, d in mathbb R $ for which the function has a limit in point 0 and in point (b) which is a separate case I have to indicate parameters for which the function is continuous in the whole field
    – MP3129
    2 days ago












  • @CameronBuie I have not a Taylor's Theorem at my university and I mustn't use facts which I did not have a lecture
    – MP3129
    2 days ago






  • 1




    @MP3129: Okay. What about L'Hôpital's Rule?
    – Cameron Buie
    2 days ago


















  • so the goal is to find the parameters such that the function defined for x negative, x positive and x = 0 is continuous ?
    – Marine Galantin
    2 days ago










  • @MP3129 Are you familiar with Taylor's Theorem?
    – Cameron Buie
    2 days ago










  • @MarineGalantin in point (a) I need to find the parameters $ a, b, c, d in mathbb R $ for which the function has a limit in point 0 and in point (b) which is a separate case I have to indicate parameters for which the function is continuous in the whole field
    – MP3129
    2 days ago












  • @CameronBuie I have not a Taylor's Theorem at my university and I mustn't use facts which I did not have a lecture
    – MP3129
    2 days ago






  • 1




    @MP3129: Okay. What about L'Hôpital's Rule?
    – Cameron Buie
    2 days ago
















so the goal is to find the parameters such that the function defined for x negative, x positive and x = 0 is continuous ?
– Marine Galantin
2 days ago




so the goal is to find the parameters such that the function defined for x negative, x positive and x = 0 is continuous ?
– Marine Galantin
2 days ago












@MP3129 Are you familiar with Taylor's Theorem?
– Cameron Buie
2 days ago




@MP3129 Are you familiar with Taylor's Theorem?
– Cameron Buie
2 days ago












@MarineGalantin in point (a) I need to find the parameters $ a, b, c, d in mathbb R $ for which the function has a limit in point 0 and in point (b) which is a separate case I have to indicate parameters for which the function is continuous in the whole field
– MP3129
2 days ago






@MarineGalantin in point (a) I need to find the parameters $ a, b, c, d in mathbb R $ for which the function has a limit in point 0 and in point (b) which is a separate case I have to indicate parameters for which the function is continuous in the whole field
– MP3129
2 days ago














@CameronBuie I have not a Taylor's Theorem at my university and I mustn't use facts which I did not have a lecture
– MP3129
2 days ago




@CameronBuie I have not a Taylor's Theorem at my university and I mustn't use facts which I did not have a lecture
– MP3129
2 days ago




1




1




@MP3129: Okay. What about L'Hôpital's Rule?
– Cameron Buie
2 days ago




@MP3129: Okay. What about L'Hôpital's Rule?
– Cameron Buie
2 days ago










1 Answer
1






active

oldest

votes


















0














The first thing to note is that as $xnearrow 0,$ the denominator $ln(1-x)$ vanishes, so the only way to avoid $bigl|f(x)bigr|to+infty$ in that case is to make sure that the numerator vanishes as well. Since $$2^{7^x-1}-ato2^{7^0-1}-a=2^{1-1}-a=2^0-a=1-a,$$ this means we must put $a=1$ to accomplish this.



If you're familiar with the result that $$lim_{tto 0}frac{sin(t)}t=1,tag{1}$$ then we can tackle the right-hand limit as follows. Since $lim_{xsearrow0}csqrt{x^2+d^2x}=0,$ then by $(1)$ we have $$lim_{xsearrow0}frac{sinleft(csqrt{x^2+d^2x}right)}{csqrt{x^2+d^2x}}=1,$$ so, since $$frac{sinleft(csqrt{x^2+d^2x}right)}{x}=frac{sinleft(csqrt{x^2+d^2x}right)}{csqrt{x^2+d^2x}}cdotfrac{csqrt{x^2+d^2x}}{x}=frac{sinleft(csqrt{x^2+d^2x}right)}{csqrt{x^2+d^2x}}cdot csqrt{1+frac{d^2}x}$$ whenever $x>0,$ then $lim_{xsearrow0}f(x)=+infty$ unless $d=0,$ in which case the limit is $c.$ Thus, for the right-hand limit to exist, we require $d=0.$ Moreover, for the two-sided limit to exist, we need $a=1,$ $d=0,$ and $c=lim_{xnearrow0}f(x),$ assuming the left-hand limit exists.



Unfortunately, I'm not familiar with any elementary ways to proceed further. Using L'Hôpital's rule allows us to find that $$begin{eqnarray}lim_{xnearrow0}frac{2^{7^x-1}-1}{ln(1-x)} &=& lim_{xnearrow0}cfrac{ln(2)ln(7)7^{x}2^{7x-1}}{-frac1{1-x}}\ &=& lim_{xnearrow0}(x-1)ln(2)ln(7)7^{x}2^{7x-1}\ &=& (0-1)ln(2)ln(7)7^{0}2^{0-1}\ &=& -frac{ln(2)ln(7)}2,end{eqnarray}$$ so for the two-sided limit to exist, we require $a=1,$ $c=frac{ln(2)ln(7)}2,$ and $d=0.$ For continuity, we additionally require $b=frac{ln(2)ln(7)}2.$






share|cite|improve this answer





















    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062828%2ffind-all-parameters-a-b-c-d-in-mathbb-r-for-which-the-function-f-mathbb-r%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0














    The first thing to note is that as $xnearrow 0,$ the denominator $ln(1-x)$ vanishes, so the only way to avoid $bigl|f(x)bigr|to+infty$ in that case is to make sure that the numerator vanishes as well. Since $$2^{7^x-1}-ato2^{7^0-1}-a=2^{1-1}-a=2^0-a=1-a,$$ this means we must put $a=1$ to accomplish this.



    If you're familiar with the result that $$lim_{tto 0}frac{sin(t)}t=1,tag{1}$$ then we can tackle the right-hand limit as follows. Since $lim_{xsearrow0}csqrt{x^2+d^2x}=0,$ then by $(1)$ we have $$lim_{xsearrow0}frac{sinleft(csqrt{x^2+d^2x}right)}{csqrt{x^2+d^2x}}=1,$$ so, since $$frac{sinleft(csqrt{x^2+d^2x}right)}{x}=frac{sinleft(csqrt{x^2+d^2x}right)}{csqrt{x^2+d^2x}}cdotfrac{csqrt{x^2+d^2x}}{x}=frac{sinleft(csqrt{x^2+d^2x}right)}{csqrt{x^2+d^2x}}cdot csqrt{1+frac{d^2}x}$$ whenever $x>0,$ then $lim_{xsearrow0}f(x)=+infty$ unless $d=0,$ in which case the limit is $c.$ Thus, for the right-hand limit to exist, we require $d=0.$ Moreover, for the two-sided limit to exist, we need $a=1,$ $d=0,$ and $c=lim_{xnearrow0}f(x),$ assuming the left-hand limit exists.



    Unfortunately, I'm not familiar with any elementary ways to proceed further. Using L'Hôpital's rule allows us to find that $$begin{eqnarray}lim_{xnearrow0}frac{2^{7^x-1}-1}{ln(1-x)} &=& lim_{xnearrow0}cfrac{ln(2)ln(7)7^{x}2^{7x-1}}{-frac1{1-x}}\ &=& lim_{xnearrow0}(x-1)ln(2)ln(7)7^{x}2^{7x-1}\ &=& (0-1)ln(2)ln(7)7^{0}2^{0-1}\ &=& -frac{ln(2)ln(7)}2,end{eqnarray}$$ so for the two-sided limit to exist, we require $a=1,$ $c=frac{ln(2)ln(7)}2,$ and $d=0.$ For continuity, we additionally require $b=frac{ln(2)ln(7)}2.$






    share|cite|improve this answer


























      0














      The first thing to note is that as $xnearrow 0,$ the denominator $ln(1-x)$ vanishes, so the only way to avoid $bigl|f(x)bigr|to+infty$ in that case is to make sure that the numerator vanishes as well. Since $$2^{7^x-1}-ato2^{7^0-1}-a=2^{1-1}-a=2^0-a=1-a,$$ this means we must put $a=1$ to accomplish this.



      If you're familiar with the result that $$lim_{tto 0}frac{sin(t)}t=1,tag{1}$$ then we can tackle the right-hand limit as follows. Since $lim_{xsearrow0}csqrt{x^2+d^2x}=0,$ then by $(1)$ we have $$lim_{xsearrow0}frac{sinleft(csqrt{x^2+d^2x}right)}{csqrt{x^2+d^2x}}=1,$$ so, since $$frac{sinleft(csqrt{x^2+d^2x}right)}{x}=frac{sinleft(csqrt{x^2+d^2x}right)}{csqrt{x^2+d^2x}}cdotfrac{csqrt{x^2+d^2x}}{x}=frac{sinleft(csqrt{x^2+d^2x}right)}{csqrt{x^2+d^2x}}cdot csqrt{1+frac{d^2}x}$$ whenever $x>0,$ then $lim_{xsearrow0}f(x)=+infty$ unless $d=0,$ in which case the limit is $c.$ Thus, for the right-hand limit to exist, we require $d=0.$ Moreover, for the two-sided limit to exist, we need $a=1,$ $d=0,$ and $c=lim_{xnearrow0}f(x),$ assuming the left-hand limit exists.



      Unfortunately, I'm not familiar with any elementary ways to proceed further. Using L'Hôpital's rule allows us to find that $$begin{eqnarray}lim_{xnearrow0}frac{2^{7^x-1}-1}{ln(1-x)} &=& lim_{xnearrow0}cfrac{ln(2)ln(7)7^{x}2^{7x-1}}{-frac1{1-x}}\ &=& lim_{xnearrow0}(x-1)ln(2)ln(7)7^{x}2^{7x-1}\ &=& (0-1)ln(2)ln(7)7^{0}2^{0-1}\ &=& -frac{ln(2)ln(7)}2,end{eqnarray}$$ so for the two-sided limit to exist, we require $a=1,$ $c=frac{ln(2)ln(7)}2,$ and $d=0.$ For continuity, we additionally require $b=frac{ln(2)ln(7)}2.$






      share|cite|improve this answer
























        0












        0








        0






        The first thing to note is that as $xnearrow 0,$ the denominator $ln(1-x)$ vanishes, so the only way to avoid $bigl|f(x)bigr|to+infty$ in that case is to make sure that the numerator vanishes as well. Since $$2^{7^x-1}-ato2^{7^0-1}-a=2^{1-1}-a=2^0-a=1-a,$$ this means we must put $a=1$ to accomplish this.



        If you're familiar with the result that $$lim_{tto 0}frac{sin(t)}t=1,tag{1}$$ then we can tackle the right-hand limit as follows. Since $lim_{xsearrow0}csqrt{x^2+d^2x}=0,$ then by $(1)$ we have $$lim_{xsearrow0}frac{sinleft(csqrt{x^2+d^2x}right)}{csqrt{x^2+d^2x}}=1,$$ so, since $$frac{sinleft(csqrt{x^2+d^2x}right)}{x}=frac{sinleft(csqrt{x^2+d^2x}right)}{csqrt{x^2+d^2x}}cdotfrac{csqrt{x^2+d^2x}}{x}=frac{sinleft(csqrt{x^2+d^2x}right)}{csqrt{x^2+d^2x}}cdot csqrt{1+frac{d^2}x}$$ whenever $x>0,$ then $lim_{xsearrow0}f(x)=+infty$ unless $d=0,$ in which case the limit is $c.$ Thus, for the right-hand limit to exist, we require $d=0.$ Moreover, for the two-sided limit to exist, we need $a=1,$ $d=0,$ and $c=lim_{xnearrow0}f(x),$ assuming the left-hand limit exists.



        Unfortunately, I'm not familiar with any elementary ways to proceed further. Using L'Hôpital's rule allows us to find that $$begin{eqnarray}lim_{xnearrow0}frac{2^{7^x-1}-1}{ln(1-x)} &=& lim_{xnearrow0}cfrac{ln(2)ln(7)7^{x}2^{7x-1}}{-frac1{1-x}}\ &=& lim_{xnearrow0}(x-1)ln(2)ln(7)7^{x}2^{7x-1}\ &=& (0-1)ln(2)ln(7)7^{0}2^{0-1}\ &=& -frac{ln(2)ln(7)}2,end{eqnarray}$$ so for the two-sided limit to exist, we require $a=1,$ $c=frac{ln(2)ln(7)}2,$ and $d=0.$ For continuity, we additionally require $b=frac{ln(2)ln(7)}2.$






        share|cite|improve this answer












        The first thing to note is that as $xnearrow 0,$ the denominator $ln(1-x)$ vanishes, so the only way to avoid $bigl|f(x)bigr|to+infty$ in that case is to make sure that the numerator vanishes as well. Since $$2^{7^x-1}-ato2^{7^0-1}-a=2^{1-1}-a=2^0-a=1-a,$$ this means we must put $a=1$ to accomplish this.



        If you're familiar with the result that $$lim_{tto 0}frac{sin(t)}t=1,tag{1}$$ then we can tackle the right-hand limit as follows. Since $lim_{xsearrow0}csqrt{x^2+d^2x}=0,$ then by $(1)$ we have $$lim_{xsearrow0}frac{sinleft(csqrt{x^2+d^2x}right)}{csqrt{x^2+d^2x}}=1,$$ so, since $$frac{sinleft(csqrt{x^2+d^2x}right)}{x}=frac{sinleft(csqrt{x^2+d^2x}right)}{csqrt{x^2+d^2x}}cdotfrac{csqrt{x^2+d^2x}}{x}=frac{sinleft(csqrt{x^2+d^2x}right)}{csqrt{x^2+d^2x}}cdot csqrt{1+frac{d^2}x}$$ whenever $x>0,$ then $lim_{xsearrow0}f(x)=+infty$ unless $d=0,$ in which case the limit is $c.$ Thus, for the right-hand limit to exist, we require $d=0.$ Moreover, for the two-sided limit to exist, we need $a=1,$ $d=0,$ and $c=lim_{xnearrow0}f(x),$ assuming the left-hand limit exists.



        Unfortunately, I'm not familiar with any elementary ways to proceed further. Using L'Hôpital's rule allows us to find that $$begin{eqnarray}lim_{xnearrow0}frac{2^{7^x-1}-1}{ln(1-x)} &=& lim_{xnearrow0}cfrac{ln(2)ln(7)7^{x}2^{7x-1}}{-frac1{1-x}}\ &=& lim_{xnearrow0}(x-1)ln(2)ln(7)7^{x}2^{7x-1}\ &=& (0-1)ln(2)ln(7)7^{0}2^{0-1}\ &=& -frac{ln(2)ln(7)}2,end{eqnarray}$$ so for the two-sided limit to exist, we require $a=1,$ $c=frac{ln(2)ln(7)}2,$ and $d=0.$ For continuity, we additionally require $b=frac{ln(2)ln(7)}2.$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        Cameron BuieCameron Buie

        85.1k771155




        85.1k771155






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062828%2ffind-all-parameters-a-b-c-d-in-mathbb-r-for-which-the-function-f-mathbb-r%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Mario Kart Wii

            Image
            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
            Read more

            The Binding of Isaac: Rebirth/Afterbirth

            Image
            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
            Read more

            What does “Dominus providebit” mean?

            Image
            5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
            Read more