Limit as $xto 0$ of a function
$begingroup$
$$lim_{xrightarrow 0} frac{e^{3x}-1}{cos left( sqrt{x}right)-1}$$
Does anyone know how to solve this limit? It gives 0/0 which normally I would use l'Hopital's rule but it seems to lead me on an endless loop which I don't know how to break out of.
limits analysis
edited Jan 8 at 14:27
amWhy
1
asked Jan 8 at 14:20
Oliver BeckOliver Beck
202
$endgroup$
add a comment |
$begingroup$
$$lim_{xrightarrow 0} frac{e^{3x}-1}{cos left( sqrt{x}right)-1}$$
Does anyone know how to solve this limit? It gives 0/0 which normally I would use l'Hopital's rule but it seems to lead me on an endless loop which I don't know how to break out of.
limits analysis
edited Jan 8 at 14:27
amWhy
1
asked Jan 8 at 14:20
Oliver BeckOliver Beck
202
$endgroup$
add a comment |
$begingroup$
$$lim_{xrightarrow 0} frac{e^{3x}-1}{cos left( sqrt{x}right)-1}$$
Does anyone know how to solve this limit? It gives 0/0 which normally I would use l'Hopital's rule but it seems to lead me on an endless loop which I don't know how to break out of.
limits analysis
edited Jan 8 at 14:27
amWhy
1
asked Jan 8 at 14:20
Oliver BeckOliver Beck
202
$endgroup$
$$lim_{xrightarrow 0} frac{e^{3x}-1}{cos left( sqrt{x}right)-1}$$
Does anyone know how to solve this limit? It gives 0/0 which normally I would use l'Hopital's rule but it seems to lead me on an endless loop which I don't know how to break out of.
limits analysis
limits analysis
edited Jan 8 at 14:27
amWhy
1
asked Jan 8 at 14:20
Oliver BeckOliver Beck
202
edited Jan 8 at 14:27
amWhy
1
asked Jan 8 at 14:20
Oliver BeckOliver Beck
202
edited Jan 8 at 14:27
amWhy
1
edited Jan 8 at 14:27
amWhy
1
edited Jan 8 at 14:27
amWhy
1
1
asked Jan 8 at 14:20
Oliver BeckOliver Beck
202
asked Jan 8 at 14:20
Oliver BeckOliver Beck
202
asked Jan 8 at 14:20
Oliver BeckOliver Beck
202
202
add a comment |
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
L'Hopital works.
$lim_{x to 0} frac{e^{3x}-1}{cos(sqrt{x})-1} = lim_{xto 0} frac{3e^{3x}}{-frac
{1}{2} frac{sin (sqrt{x})}{sqrt{x}}}= frac{3}{-frac{1}{2}*1} = -6.
$
answered Jan 8 at 14:30
SchlubbidubbiSchlubbidubbi
914
$endgroup$
add a comment |
$begingroup$
Let $sqrt x=2h,x=4h^2$
$$dfrac{e^{12h^2}-1}{cos2h-1}=-dfrac{e^{12h^2}-1}{12h^2}left(dfrac h{sin h}right)^2dfrac{12}2$$
answered Jan 8 at 14:35
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
add a comment |
$begingroup$
Hint:
$$lim_{xto0}{e^{3x}-1overcos(sqrt x)-1}=lim_{uto0^+}{e^{3u^2}-1overcos u-1}$$
One or two rounds of L'Hopital now does the trick.
answered Jan 8 at 14:37
Barry CipraBarry Cipra
59.4k653125
$endgroup$
add a comment |
$begingroup$
Hint:
- When $x$ is small, $e^x approx 1+x$ and $cos x approx 1-frac{x^2}2$.
answered Jan 8 at 14:39
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
add a comment |
$begingroup$
Use series expansion for $e^{3x}$ and $cossqrt{x}.$
$displaystylelim_{xrightarrow 0} dfrac{e^{3x}-1}{cos left( sqrt{x}right)-1}=displaystylelim_{xrightarrow 0} dfrac{left(1+3x+frac{(3x)^2}{2!}+cdotsright)-1}{left(1-frac{x}{2!}+frac{x^2}{4!}+cdotsright)-1}=displaystylelim_{xrightarrow 0} dfrac{3x+O(x^2)}{-frac{x}{2}+O(x^2)}=displaystylelim_{xrightarrow 0} dfrac{3x}{-frac{x}{2}}=6$
answered Jan 8 at 14:43
Arjun BanerjeeArjun Banerjee
43610
$endgroup$
add a comment |
$begingroup$
You can solve the limit without L’Hôpital as well, such as via the series expansion:
$$cos x = 1-frac{x^2}{2!}+frac{x^4}{4!}-… implies cos x sim 1-frac{x^2}{2}; quad xto 0$$
$$e^x = 1+x+frac{x^2}{2!}+… implies e^x sim 1+x; quad xto 0$$
Hence, for $x to 0$, $cossqrt{x} sim 1-frac{x}{2}$ and $e^{3x} sim 1+3x$. Hence, the limit simplifies to
$$frac{1+3x-1}{1-frac{x}{2}-1} = frac{3x}{-frac{x}{2}} = -6$$
answered Jan 8 at 14:50
KM101KM101
5,9611523
$endgroup$
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
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oldest
votes
$begingroup$
L'Hopital works.
$lim_{x to 0} frac{e^{3x}-1}{cos(sqrt{x})-1} = lim_{xto 0} frac{3e^{3x}}{-frac
{1}{2} frac{sin (sqrt{x})}{sqrt{x}}}= frac{3}{-frac{1}{2}*1} = -6.
$
answered Jan 8 at 14:30
SchlubbidubbiSchlubbidubbi
914
$endgroup$
add a comment |
$begingroup$
L'Hopital works.
$lim_{x to 0} frac{e^{3x}-1}{cos(sqrt{x})-1} = lim_{xto 0} frac{3e^{3x}}{-frac
{1}{2} frac{sin (sqrt{x})}{sqrt{x}}}= frac{3}{-frac{1}{2}*1} = -6.
$
answered Jan 8 at 14:30
SchlubbidubbiSchlubbidubbi
914
$endgroup$
add a comment |
$begingroup$
L'Hopital works.
$lim_{x to 0} frac{e^{3x}-1}{cos(sqrt{x})-1} = lim_{xto 0} frac{3e^{3x}}{-frac
{1}{2} frac{sin (sqrt{x})}{sqrt{x}}}= frac{3}{-frac{1}{2}*1} = -6.
$
answered Jan 8 at 14:30
SchlubbidubbiSchlubbidubbi
914
$endgroup$
L'Hopital works.
$lim_{x to 0} frac{e^{3x}-1}{cos(sqrt{x})-1} = lim_{xto 0} frac{3e^{3x}}{-frac
{1}{2} frac{sin (sqrt{x})}{sqrt{x}}}= frac{3}{-frac{1}{2}*1} = -6.
$
answered Jan 8 at 14:30
SchlubbidubbiSchlubbidubbi
914
answered Jan 8 at 14:30
SchlubbidubbiSchlubbidubbi
914
answered Jan 8 at 14:30
SchlubbidubbiSchlubbidubbi
914
answered Jan 8 at 14:30
SchlubbidubbiSchlubbidubbi
914
914
add a comment |
add a comment |
$begingroup$
Let $sqrt x=2h,x=4h^2$
$$dfrac{e^{12h^2}-1}{cos2h-1}=-dfrac{e^{12h^2}-1}{12h^2}left(dfrac h{sin h}right)^2dfrac{12}2$$
answered Jan 8 at 14:35
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
add a comment |
$begingroup$
Let $sqrt x=2h,x=4h^2$
$$dfrac{e^{12h^2}-1}{cos2h-1}=-dfrac{e^{12h^2}-1}{12h^2}left(dfrac h{sin h}right)^2dfrac{12}2$$
answered Jan 8 at 14:35
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
add a comment |
$begingroup$
Let $sqrt x=2h,x=4h^2$
$$dfrac{e^{12h^2}-1}{cos2h-1}=-dfrac{e^{12h^2}-1}{12h^2}left(dfrac h{sin h}right)^2dfrac{12}2$$
answered Jan 8 at 14:35
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
Let $sqrt x=2h,x=4h^2$
$$dfrac{e^{12h^2}-1}{cos2h-1}=-dfrac{e^{12h^2}-1}{12h^2}left(dfrac h{sin h}right)^2dfrac{12}2$$
answered Jan 8 at 14:35
lab bhattacharjeelab bhattacharjee
224k15156274
answered Jan 8 at 14:35
lab bhattacharjeelab bhattacharjee
224k15156274
answered Jan 8 at 14:35
lab bhattacharjeelab bhattacharjee
224k15156274
answered Jan 8 at 14:35
lab bhattacharjeelab bhattacharjee
224k15156274
224k15156274
add a comment |
add a comment |
$begingroup$
Hint:
$$lim_{xto0}{e^{3x}-1overcos(sqrt x)-1}=lim_{uto0^+}{e^{3u^2}-1overcos u-1}$$
One or two rounds of L'Hopital now does the trick.
answered Jan 8 at 14:37
Barry CipraBarry Cipra
59.4k653125
$endgroup$
add a comment |
$begingroup$
Hint:
$$lim_{xto0}{e^{3x}-1overcos(sqrt x)-1}=lim_{uto0^+}{e^{3u^2}-1overcos u-1}$$
One or two rounds of L'Hopital now does the trick.
answered Jan 8 at 14:37
Barry CipraBarry Cipra
59.4k653125
$endgroup$
add a comment |
$begingroup$
Hint:
$$lim_{xto0}{e^{3x}-1overcos(sqrt x)-1}=lim_{uto0^+}{e^{3u^2}-1overcos u-1}$$
One or two rounds of L'Hopital now does the trick.
answered Jan 8 at 14:37
Barry CipraBarry Cipra
59.4k653125
$endgroup$
Hint:
$$lim_{xto0}{e^{3x}-1overcos(sqrt x)-1}=lim_{uto0^+}{e^{3u^2}-1overcos u-1}$$
One or two rounds of L'Hopital now does the trick.
answered Jan 8 at 14:37
Barry CipraBarry Cipra
59.4k653125
answered Jan 8 at 14:37
Barry CipraBarry Cipra
59.4k653125
answered Jan 8 at 14:37
Barry CipraBarry Cipra
59.4k653125
answered Jan 8 at 14:37
Barry CipraBarry Cipra
59.4k653125
59.4k653125
add a comment |
add a comment |
$begingroup$
Hint:
- When $x$ is small, $e^x approx 1+x$ and $cos x approx 1-frac{x^2}2$.
answered Jan 8 at 14:39
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
add a comment |
$begingroup$
Hint:
- When $x$ is small, $e^x approx 1+x$ and $cos x approx 1-frac{x^2}2$.
answered Jan 8 at 14:39
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
add a comment |
$begingroup$
Hint:
- When $x$ is small, $e^x approx 1+x$ and $cos x approx 1-frac{x^2}2$.
answered Jan 8 at 14:39
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
Hint:
- When $x$ is small, $e^x approx 1+x$ and $cos x approx 1-frac{x^2}2$.
answered Jan 8 at 14:39
Siong Thye GohSiong Thye Goh
100k1465117
answered Jan 8 at 14:39
Siong Thye GohSiong Thye Goh
100k1465117
answered Jan 8 at 14:39
Siong Thye GohSiong Thye Goh
100k1465117
answered Jan 8 at 14:39
Siong Thye GohSiong Thye Goh
100k1465117
100k1465117
add a comment |
add a comment |
$begingroup$
Use series expansion for $e^{3x}$ and $cossqrt{x}.$
$displaystylelim_{xrightarrow 0} dfrac{e^{3x}-1}{cos left( sqrt{x}right)-1}=displaystylelim_{xrightarrow 0} dfrac{left(1+3x+frac{(3x)^2}{2!}+cdotsright)-1}{left(1-frac{x}{2!}+frac{x^2}{4!}+cdotsright)-1}=displaystylelim_{xrightarrow 0} dfrac{3x+O(x^2)}{-frac{x}{2}+O(x^2)}=displaystylelim_{xrightarrow 0} dfrac{3x}{-frac{x}{2}}=6$
answered Jan 8 at 14:43
Arjun BanerjeeArjun Banerjee
43610
$endgroup$
add a comment |
$begingroup$
Use series expansion for $e^{3x}$ and $cossqrt{x}.$
$displaystylelim_{xrightarrow 0} dfrac{e^{3x}-1}{cos left( sqrt{x}right)-1}=displaystylelim_{xrightarrow 0} dfrac{left(1+3x+frac{(3x)^2}{2!}+cdotsright)-1}{left(1-frac{x}{2!}+frac{x^2}{4!}+cdotsright)-1}=displaystylelim_{xrightarrow 0} dfrac{3x+O(x^2)}{-frac{x}{2}+O(x^2)}=displaystylelim_{xrightarrow 0} dfrac{3x}{-frac{x}{2}}=6$
answered Jan 8 at 14:43
Arjun BanerjeeArjun Banerjee
43610
$endgroup$
add a comment |
$begingroup$
Use series expansion for $e^{3x}$ and $cossqrt{x}.$
$displaystylelim_{xrightarrow 0} dfrac{e^{3x}-1}{cos left( sqrt{x}right)-1}=displaystylelim_{xrightarrow 0} dfrac{left(1+3x+frac{(3x)^2}{2!}+cdotsright)-1}{left(1-frac{x}{2!}+frac{x^2}{4!}+cdotsright)-1}=displaystylelim_{xrightarrow 0} dfrac{3x+O(x^2)}{-frac{x}{2}+O(x^2)}=displaystylelim_{xrightarrow 0} dfrac{3x}{-frac{x}{2}}=6$
answered Jan 8 at 14:43
Arjun BanerjeeArjun Banerjee
43610
$endgroup$
Use series expansion for $e^{3x}$ and $cossqrt{x}.$
$displaystylelim_{xrightarrow 0} dfrac{e^{3x}-1}{cos left( sqrt{x}right)-1}=displaystylelim_{xrightarrow 0} dfrac{left(1+3x+frac{(3x)^2}{2!}+cdotsright)-1}{left(1-frac{x}{2!}+frac{x^2}{4!}+cdotsright)-1}=displaystylelim_{xrightarrow 0} dfrac{3x+O(x^2)}{-frac{x}{2}+O(x^2)}=displaystylelim_{xrightarrow 0} dfrac{3x}{-frac{x}{2}}=6$
answered Jan 8 at 14:43
Arjun BanerjeeArjun Banerjee
43610
answered Jan 8 at 14:43
Arjun BanerjeeArjun Banerjee
43610
answered Jan 8 at 14:43
Arjun BanerjeeArjun Banerjee
43610
answered Jan 8 at 14:43
Arjun BanerjeeArjun Banerjee
43610
43610
add a comment |
add a comment |
$begingroup$
You can solve the limit without L’Hôpital as well, such as via the series expansion:
$$cos x = 1-frac{x^2}{2!}+frac{x^4}{4!}-… implies cos x sim 1-frac{x^2}{2}; quad xto 0$$
$$e^x = 1+x+frac{x^2}{2!}+… implies e^x sim 1+x; quad xto 0$$
Hence, for $x to 0$, $cossqrt{x} sim 1-frac{x}{2}$ and $e^{3x} sim 1+3x$. Hence, the limit simplifies to
$$frac{1+3x-1}{1-frac{x}{2}-1} = frac{3x}{-frac{x}{2}} = -6$$
answered Jan 8 at 14:50
KM101KM101
5,9611523
$endgroup$
add a comment |
$begingroup$
You can solve the limit without L’Hôpital as well, such as via the series expansion:
$$cos x = 1-frac{x^2}{2!}+frac{x^4}{4!}-… implies cos x sim 1-frac{x^2}{2}; quad xto 0$$
$$e^x = 1+x+frac{x^2}{2!}+… implies e^x sim 1+x; quad xto 0$$
Hence, for $x to 0$, $cossqrt{x} sim 1-frac{x}{2}$ and $e^{3x} sim 1+3x$. Hence, the limit simplifies to
$$frac{1+3x-1}{1-frac{x}{2}-1} = frac{3x}{-frac{x}{2}} = -6$$
answered Jan 8 at 14:50
KM101KM101
5,9611523
$endgroup$
add a comment |
$begingroup$
You can solve the limit without L’Hôpital as well, such as via the series expansion:
$$cos x = 1-frac{x^2}{2!}+frac{x^4}{4!}-… implies cos x sim 1-frac{x^2}{2}; quad xto 0$$
$$e^x = 1+x+frac{x^2}{2!}+… implies e^x sim 1+x; quad xto 0$$
Hence, for $x to 0$, $cossqrt{x} sim 1-frac{x}{2}$ and $e^{3x} sim 1+3x$. Hence, the limit simplifies to
$$frac{1+3x-1}{1-frac{x}{2}-1} = frac{3x}{-frac{x}{2}} = -6$$
answered Jan 8 at 14:50
KM101KM101
5,9611523
$endgroup$
You can solve the limit without L’Hôpital as well, such as via the series expansion:
$$cos x = 1-frac{x^2}{2!}+frac{x^4}{4!}-… implies cos x sim 1-frac{x^2}{2}; quad xto 0$$
$$e^x = 1+x+frac{x^2}{2!}+… implies e^x sim 1+x; quad xto 0$$
Hence, for $x to 0$, $cossqrt{x} sim 1-frac{x}{2}$ and $e^{3x} sim 1+3x$. Hence, the limit simplifies to
$$frac{1+3x-1}{1-frac{x}{2}-1} = frac{3x}{-frac{x}{2}} = -6$$
answered Jan 8 at 14:50
KM101KM101
5,9611523
answered Jan 8 at 14:50
KM101KM101
5,9611523
answered Jan 8 at 14:50
KM101KM101
5,9611523
answered Jan 8 at 14:50
KM101KM101
5,9611523
5,9611523
add a comment |
add a comment |
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