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How can I show that $\lim_{x\to 0}\frac{\ln(\sin x+1)}{\sin x} = 1$?

I think that this function is continuous (maybe even uniform continuous) in all $R$ but $0$.

But because it is undefined at $x=0$ I'm not sure what I can do here.

Is there a known way for finding limits in situations like this? I thought of using the squeezing theorem somehow but couldn't find a way.

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  • $\begingroup$ You can do a Taylor expansion on $\ln (1+u) = u+o(u^2)$ and then say $u = \sin x$ $\endgroup$ – Doug M Jan 9 '17 at 21:51
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Since $\lim_{x\rightarrow 0} \ln (\sin x +1)=0$ and $\lim_{x \rightarrow 0} \sin(x) = 0$,

By L'Hôpital's rule,

$$\lim_{x\rightarrow 0}\frac{\ln (\sin x + 1)}{\sin x}=\lim_{x\rightarrow 0}\frac{\cos x}{(\sin x +1)\cos x}=1$$

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  • $\begingroup$ You may need to add $\ln(\sin x+1)\to 0$ and $\sin x\to 0$ as $x\to 0$. $\endgroup$ – user236182 Jan 9 '17 at 22:15
  • $\begingroup$ thanks. Sure, those are the sufficient conditions to use L'Hôpital's rule. $\endgroup$ – Siong Thye Goh Jan 9 '17 at 22:18
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Call $\sin x=u$ and then rewrite the limit as $\ln\left ( \lim_{u\to0}(1+u)^{1/u}\right )$ and that inside limit is just $e$ so $\ln e=1$

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Since plugging $0$ into the limit would result in an expressions of the form $\frac{0}{0}$, we can apply L'Hôpital's rule to the expression.

\begin{align} \lim\limits_{x\rightarrow0}\frac{\ln(\sin x + 1)}{\sin(x)} &= \lim\limits_{x\rightarrow0}\frac{1}{\sin x + 1}\cdot\frac{\cos x}{\cos x}\\ &=\lim\limits_{x\rightarrow0}\frac{1}{\sin x + 1}\\ &= 1 \end{align}

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Substituting $\sin x=t$,

$$\lim_{x\to 0}\frac{\ln(\sin x+1)}{\sin x} = \lim_{t\to 0}\frac{\ln(t+1)}t $$ which is known to be $1$.

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Taylor expand twice, first order actually suffices.

For $x\sim 0$ we have $$\sin x\sim 0$$ and for $x\sim 0$ we have $$ \ln(1+x)\sim x $$ Making your limit pretty easy $$ \lim_{x\to 0}\frac{\ln(\sin x+1)}{\sin x}=\lim_{x\to 0}\frac{x}{ x}=1 $$

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Although somewhat cumbersome, consider the following argument:

First note that for $x \to 0$ we have $\sin(x) \to 0$. The series expansion of the natural logarithm around $1$ is given by $\log(1 + x) = x + \mathcal{O}(x^2)$. For sin we have $\sin(x) = x + \mathcal{O}(x^3)$ Thus

\begin{align} \lim_{x \to 0} \frac{\log(\sin(x) + 1)}{\sin(x)} &= \lim_{x \to 0} \frac{x + \mathcal{O}(x^2)}{x + \mathcal{O}(x^3)} \\ &= \lim_{x \to 0} \frac{1 + \mathcal{O}(x)}{1 + \mathcal{O}(x^2)} \\ &= 1 \end{align}

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