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I'm stuck in this limit:

$$\lim_{x\to 0}\frac{x(e^x-1)}{\cos x-1}$$

I tried to solve it using special limits, so:

$$\lim_{x\to 0}\frac{x(e^x-1)}{\cos x-1}=$$

$$=\lim_{x\to 0}(e^x-1)\frac{x(\cos x+1)}{(\cos x-1)(\cos x+1)}=$$

$$=\lim_{x\to 0}-\frac{x}{\sin^2 x}(e^x-1)(\cos x+1)=$$

$$=\lim_{x\to 0}-\frac{x}{\sin x}\cdot\frac{e^x-1}{\sin x}\cdot(\cos x+1)$$

I know that $\lim_{x\to 0}\left(\frac{x}{\sin x}\right)=1$

I know also the special limit $\lim_{x\to 0}\left(\frac{e^x-1}{x}\right)=1$

My question is: can I intend $\lim_{x\to 0}\left(\frac{e^x-1}{\sin x}\right)$ equal to $\lim_{x\to 0}\left(\frac{e^x-1}{x}\right)$ because $x$ and $\sin x$ tend both to $0$ for ${x\to 0}$?

If not, how can I solve my limit, without series or something difficult?

Thank you!

Michael Hardy
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xdola
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3 Answers3

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Yes, you can says that $\lim_{x\to 0}\dfrac{e^x-1}{\sin x}$ equal to $\lim_{x\to 0}\dfrac{e^x-1}{x}$, but not just because $x$ and $\sin x$ tend both to $0$ for ${x\to 0}$, this is not enough.

You can write $$\lim_{x\to 0}\dfrac{e^x-1}{\sin x}=\lim_{x\to 0}\dfrac{e^x-1}{\sin x}\frac xx=\lim_{x\to 0}\dfrac{e^x-1}x\frac x{\sin x}=\lim_{x\to 0}\dfrac{e^x-1}x\lim_{x\to 0}\frac x{\sin x}.$$

Now you get the equivalence from

$$\lim_{x\to 0}\frac x{\sin x}=1.$$

This is justified by the property that the limit of a product is the product of the limits, provided they exist.

2

$$\lim _{ x\rightarrow 0 }{ \frac { x\left( { e }^{ x }-1 \right) }{ \cos { x } -1 } = } \lim _{ x\rightarrow 0 }{ \frac { { x }^{ 2 }\left( { e }^{ x }-1 \right) }{ x\left( \cos { x } -1 \right) } = } \lim _{ x\rightarrow 0 }{ \frac { \left( { e }^{ x }-1 \right) }{ x } \lim _{ x\rightarrow 0 }{ \frac { { x }^{ 2 } }{ -2\sin ^{ 2 }{ \frac { x }{ 2 } } } = } } \\ =\lim _{ x\rightarrow 0 }{ \frac { \left( { e }^{ x }-1 \right) }{ x } \lim _{ x\rightarrow 0 }{ \frac { { x }^{ 2 } }{ -2\sin ^{ 2 }{ \frac { x }{ 2 } } } = } } -\frac { 1 }{ 2 } \lim _{ x\rightarrow 0 }{ \frac { \left( { e }^{ x }-1 \right) }{ x } \lim _{ x\rightarrow 0 }{ \frac { 4 }{ { \left( \frac { \sin { \frac { x }{ 2 } } }{ \frac { x }{ 2 } } \right) }^{ 2 } } = } } -2$$

haqnatural
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If $\lim_{x\to0}f(x)=0$ and $\lim_{x\to0}g(x)=0$ than $\lim_{x\to0}\frac{f(x)}{g(x)}=\lim_{x\to0}\frac{f'(x)}{g'(x)}$. This is the L'Hospital rule.

Use that for your limit: $$\lim_{x\to0}\frac{x(e^x-1)}{\cos x-1}$$ $$f(x)=x(e^x-1)$$ $$f'(x)=(x+1)e^x-1$$ $$g(x)=\cos x-1$$ $$g'(x)=-\sin x$$ $$\lim_{x\to0}\frac{f(x)}{g(x)}=\lim_{x\to0}\frac{f'(x)}{g'(x)}$$ $$\lim_{x\to0}\frac{x(e^x-1)}{\cos x-1}=\lim_{x\to0}\frac{(x+1)e^x-1}{-\sin x}$$ Use the rule again: $$f''(x)=(x+2)e^x$$ $$g''(x)=-\cos x$$ $$\lim_{x\to0}\frac{x(e^x-1)}{\cos x-1}=\lim_{x\to0}\frac{(x+1)e^x-1}{-\sin x}=\lim_{x\to0}\frac{(x+2)e^x}{-\cos x}$$ $$\lim_{x\to0}\frac{(x+2)e^x}{-\cos x}=\frac{(0+2)e^0}{-\cos 0}=-2$$

Wyctus
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