Find the value of $f(x)$ with proof.












0












$begingroup$


The equation is:



$$lim_{x to 2} frac{f(x) - 1}{x - 2} = 3$$



Suppose f(x) is a function. Find (with proof) the value of lim x→2 f(x), or conclude that this limit diverges.



Our professor wrote in a hint stating: Use Arithmetic of Limits and contradiction. Why can't $lim_{xto2}
f(x) = 0$, for
instance?



It is a question from a previous midterm and noone ive spoken to has solved it,so i wanna ask you guys.






real-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    Didn't you forget a "little" detail? What is $;f(x);$ , anyway?!
    $endgroup$
    – DonAntonio
    Jan 12 at 18:12










  • $begingroup$
    Its a function on x.
    $endgroup$
    – user569685
    Jan 12 at 18:16










  • $begingroup$
    Then perhaps the actual question is "what is the value of $;f(2);$ ..."?? Because the value of $;f(x);$ cannot be known from the given info.
    $endgroup$
    – DonAntonio
    Jan 12 at 18:16








  • 1




    $begingroup$
    @ThomasShelby I saw it...read my comment below such answer. Apparently the OP meant to ask what is the value of $;f(2);$ ...
    $endgroup$
    – DonAntonio
    Jan 12 at 18:17












  • $begingroup$
    Yes, you're right. I didn't read the question properly.@DonAntonio
    $endgroup$
    – Thomas Shelby
    Jan 12 at 18:19


















0












$begingroup$


The equation is:



$$lim_{x to 2} frac{f(x) - 1}{x - 2} = 3$$



Suppose f(x) is a function. Find (with proof) the value of lim x→2 f(x), or conclude that this limit diverges.



Our professor wrote in a hint stating: Use Arithmetic of Limits and contradiction. Why can't $lim_{xto2}
f(x) = 0$, for
instance?



It is a question from a previous midterm and noone ive spoken to has solved it,so i wanna ask you guys.






real-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    Didn't you forget a "little" detail? What is $;f(x);$ , anyway?!
    $endgroup$
    – DonAntonio
    Jan 12 at 18:12










  • $begingroup$
    Its a function on x.
    $endgroup$
    – user569685
    Jan 12 at 18:16










  • $begingroup$
    Then perhaps the actual question is "what is the value of $;f(2);$ ..."?? Because the value of $;f(x);$ cannot be known from the given info.
    $endgroup$
    – DonAntonio
    Jan 12 at 18:16








  • 1




    $begingroup$
    @ThomasShelby I saw it...read my comment below such answer. Apparently the OP meant to ask what is the value of $;f(2);$ ...
    $endgroup$
    – DonAntonio
    Jan 12 at 18:17












  • $begingroup$
    Yes, you're right. I didn't read the question properly.@DonAntonio
    $endgroup$
    – Thomas Shelby
    Jan 12 at 18:19
















0












0








0





$begingroup$


The equation is:



$$lim_{x to 2} frac{f(x) - 1}{x - 2} = 3$$



Suppose f(x) is a function. Find (with proof) the value of lim x→2 f(x), or conclude that this limit diverges.



Our professor wrote in a hint stating: Use Arithmetic of Limits and contradiction. Why can't $lim_{xto2}
f(x) = 0$, for
instance?



It is a question from a previous midterm and noone ive spoken to has solved it,so i wanna ask you guys.






real-analysis







share|cite|improve this question











$endgroup$




The equation is:



$$lim_{x to 2} frac{f(x) - 1}{x - 2} = 3$$



Suppose f(x) is a function. Find (with proof) the value of lim x→2 f(x), or conclude that this limit diverges.



Our professor wrote in a hint stating: Use Arithmetic of Limits and contradiction. Why can't $lim_{xto2}
f(x) = 0$, for
instance?



It is a question from a previous midterm and noone ive spoken to has solved it,so i wanna ask you guys.






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 Jan 12 at 18:26







user569685

















asked Jan 12 at 18:05









user569685user569685

102




102












  • $begingroup$
    Didn't you forget a "little" detail? What is $;f(x);$ , anyway?!
    $endgroup$
    – DonAntonio
    Jan 12 at 18:12










  • $begingroup$
    Its a function on x.
    $endgroup$
    – user569685
    Jan 12 at 18:16










  • $begingroup$
    Then perhaps the actual question is "what is the value of $;f(2);$ ..."?? Because the value of $;f(x);$ cannot be known from the given info.
    $endgroup$
    – DonAntonio
    Jan 12 at 18:16








  • 1




    $begingroup$
    @ThomasShelby I saw it...read my comment below such answer. Apparently the OP meant to ask what is the value of $;f(2);$ ...
    $endgroup$
    – DonAntonio
    Jan 12 at 18:17












  • $begingroup$
    Yes, you're right. I didn't read the question properly.@DonAntonio
    $endgroup$
    – Thomas Shelby
    Jan 12 at 18:19




















  • $begingroup$
    Didn't you forget a "little" detail? What is $;f(x);$ , anyway?!
    $endgroup$
    – DonAntonio
    Jan 12 at 18:12










  • $begingroup$
    Its a function on x.
    $endgroup$
    – user569685
    Jan 12 at 18:16










  • $begingroup$
    Then perhaps the actual question is "what is the value of $;f(2);$ ..."?? Because the value of $;f(x);$ cannot be known from the given info.
    $endgroup$
    – DonAntonio
    Jan 12 at 18:16








  • 1




    $begingroup$
    @ThomasShelby I saw it...read my comment below such answer. Apparently the OP meant to ask what is the value of $;f(2);$ ...
    $endgroup$
    – DonAntonio
    Jan 12 at 18:17












  • $begingroup$
    Yes, you're right. I didn't read the question properly.@DonAntonio
    $endgroup$
    – Thomas Shelby
    Jan 12 at 18:19


















$begingroup$
Didn't you forget a "little" detail? What is $;f(x);$ , anyway?!
$endgroup$
– DonAntonio
Jan 12 at 18:12




$begingroup$
Didn't you forget a "little" detail? What is $;f(x);$ , anyway?!
$endgroup$
– DonAntonio
Jan 12 at 18:12












$begingroup$
Its a function on x.
$endgroup$
– user569685
Jan 12 at 18:16




$begingroup$
Its a function on x.
$endgroup$
– user569685
Jan 12 at 18:16












$begingroup$
Then perhaps the actual question is "what is the value of $;f(2);$ ..."?? Because the value of $;f(x);$ cannot be known from the given info.
$endgroup$
– DonAntonio
Jan 12 at 18:16






$begingroup$
Then perhaps the actual question is "what is the value of $;f(2);$ ..."?? Because the value of $;f(x);$ cannot be known from the given info.
$endgroup$
– DonAntonio
Jan 12 at 18:16






1




1




$begingroup$
@ThomasShelby I saw it...read my comment below such answer. Apparently the OP meant to ask what is the value of $;f(2);$ ...
$endgroup$
– DonAntonio
Jan 12 at 18:17






$begingroup$
@ThomasShelby I saw it...read my comment below such answer. Apparently the OP meant to ask what is the value of $;f(2);$ ...
$endgroup$
– DonAntonio
Jan 12 at 18:17














$begingroup$
Yes, you're right. I didn't read the question properly.@DonAntonio
$endgroup$
– Thomas Shelby
Jan 12 at 18:19






$begingroup$
Yes, you're right. I didn't read the question properly.@DonAntonio
$endgroup$
– Thomas Shelby
Jan 12 at 18:19












1 Answer
1






active

oldest

votes


















3












$begingroup$

Hint



We know that
$$
lim_{xto 2}(f(x)-1)=left(lim_{xto 2}frac{f(x)-1}{x-2}right)timeslim_{xto2}(x-2)tag{0}
$$
by the product rule for limits and
$$
lim_{xto2}f(x)=lim_{xto 2}(f(x)-1)+lim_{xto2}1tag{1}.
$$
You have enough information to compute all the limits.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes, the limits can be computed, apparently (if the OP did give all the relevant info...), yet the question seems to be what is the value of $;f;$ ...perhaps when $;x=2;$ or something like that.
    $endgroup$
    – DonAntonio
    Jan 12 at 18:15










  • $begingroup$
    @user569685 Great! Why then in the world didn't you write this in your question?!
    $endgroup$
    – DonAntonio
    Jan 12 at 18:25













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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Hint



We know that
$$
lim_{xto 2}(f(x)-1)=left(lim_{xto 2}frac{f(x)-1}{x-2}right)timeslim_{xto2}(x-2)tag{0}
$$
by the product rule for limits and
$$
lim_{xto2}f(x)=lim_{xto 2}(f(x)-1)+lim_{xto2}1tag{1}.
$$
You have enough information to compute all the limits.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes, the limits can be computed, apparently (if the OP did give all the relevant info...), yet the question seems to be what is the value of $;f;$ ...perhaps when $;x=2;$ or something like that.
    $endgroup$
    – DonAntonio
    Jan 12 at 18:15










  • $begingroup$
    @user569685 Great! Why then in the world didn't you write this in your question?!
    $endgroup$
    – DonAntonio
    Jan 12 at 18:25


















3












$begingroup$

Hint



We know that
$$
lim_{xto 2}(f(x)-1)=left(lim_{xto 2}frac{f(x)-1}{x-2}right)timeslim_{xto2}(x-2)tag{0}
$$
by the product rule for limits and
$$
lim_{xto2}f(x)=lim_{xto 2}(f(x)-1)+lim_{xto2}1tag{1}.
$$
You have enough information to compute all the limits.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes, the limits can be computed, apparently (if the OP did give all the relevant info...), yet the question seems to be what is the value of $;f;$ ...perhaps when $;x=2;$ or something like that.
    $endgroup$
    – DonAntonio
    Jan 12 at 18:15










  • $begingroup$
    @user569685 Great! Why then in the world didn't you write this in your question?!
    $endgroup$
    – DonAntonio
    Jan 12 at 18:25
















3












3








3





$begingroup$

Hint



We know that
$$
lim_{xto 2}(f(x)-1)=left(lim_{xto 2}frac{f(x)-1}{x-2}right)timeslim_{xto2}(x-2)tag{0}
$$
by the product rule for limits and
$$
lim_{xto2}f(x)=lim_{xto 2}(f(x)-1)+lim_{xto2}1tag{1}.
$$
You have enough information to compute all the limits.






share|cite|improve this answer









$endgroup$



Hint



We know that
$$
lim_{xto 2}(f(x)-1)=left(lim_{xto 2}frac{f(x)-1}{x-2}right)timeslim_{xto2}(x-2)tag{0}
$$
by the product rule for limits and
$$
lim_{xto2}f(x)=lim_{xto 2}(f(x)-1)+lim_{xto2}1tag{1}.
$$
You have enough information to compute all the limits.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 12 at 18:08









Foobaz JohnFoobaz John

21.9k41352




21.9k41352












  • $begingroup$
    Yes, the limits can be computed, apparently (if the OP did give all the relevant info...), yet the question seems to be what is the value of $;f;$ ...perhaps when $;x=2;$ or something like that.
    $endgroup$
    – DonAntonio
    Jan 12 at 18:15










  • $begingroup$
    @user569685 Great! Why then in the world didn't you write this in your question?!
    $endgroup$
    – DonAntonio
    Jan 12 at 18:25




















  • $begingroup$
    Yes, the limits can be computed, apparently (if the OP did give all the relevant info...), yet the question seems to be what is the value of $;f;$ ...perhaps when $;x=2;$ or something like that.
    $endgroup$
    – DonAntonio
    Jan 12 at 18:15










  • $begingroup$
    @user569685 Great! Why then in the world didn't you write this in your question?!
    $endgroup$
    – DonAntonio
    Jan 12 at 18:25


















$begingroup$
Yes, the limits can be computed, apparently (if the OP did give all the relevant info...), yet the question seems to be what is the value of $;f;$ ...perhaps when $;x=2;$ or something like that.
$endgroup$
– DonAntonio
Jan 12 at 18:15




$begingroup$
Yes, the limits can be computed, apparently (if the OP did give all the relevant info...), yet the question seems to be what is the value of $;f;$ ...perhaps when $;x=2;$ or something like that.
$endgroup$
– DonAntonio
Jan 12 at 18:15












$begingroup$
@user569685 Great! Why then in the world didn't you write this in your question?!
$endgroup$
– DonAntonio
Jan 12 at 18:25






$begingroup$
@user569685 Great! Why then in the world didn't you write this in your question?!
$endgroup$
– DonAntonio
Jan 12 at 18:25




















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