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What is the limit of the following expression :

$$\lim_{x \rightarrow 0 } (1+\sin x) ^{\frac{1}{x}}$$

I tried doing the following :

$$\lim_{x \rightarrow 0 } (1+\sin x) ^{{\frac{1}{\sin x}}{\frac{\sin x }{x}}}$$

Now I know the formula $\lim_{x \rightarrow 0} (1+x)^{\frac{1}{x}} = e$ . I can do that in the above expression for $\sin x \rightarrow 0$ .

But I'm not sure if the limit $\lim_{x \rightarrow 0}$ can propagate up-to the power. If not, if I take $ln$ in both sides, can then the limit propagate into the expression within the $ln$ operator ?

Bernard
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Entrepreneur
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5 Answers5

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It is good to take $\log$ when you are solving $1^{\infty}$ limits $$y=\lim_{x \rightarrow 0 } (1+\sin x) ^{\frac{1}{x}}$$ $$\log y=\lim_{x \rightarrow 0 } \frac{\log ({1+\sin x})}{x}$$ now use LHopital rule on the right hand side limit it becomes

$$\log y=\lim_{x \rightarrow 0 } \frac{\cos x}{1+\sin x}=1$$

$$\log y=1$$ $$y=e^1$$

Deepesh Meena
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You basically figured it out already. As all mentioned functions are continuous, we have:

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

I will leave to part to argue, that $\lim_{x\to 0}((1+\sin x)^{\frac{1}{\sin x}})=e$ to you though.

oatdrinkisnice
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Start by rewriting by following the exponential rule: \begin{equation} \lim_{x\to 0} (1+\sin(x))^\frac{1}{x} =\lim_{x\to 0} e^{\frac{\ln(1+\sin(x))}{x}}. \end{equation}

Using the chain rule, let everything in the exponent become the inner function. Then, simplify the inner function using L'Hopital's Rule: \begin{equation} \lim_{x\to 0} \frac{\ln(1+\sin(x))}{x} = \lim_{x\to0}\frac{\cos(x)}{sin(x)+1} \\ = \frac{\cos(0)}{sin(0)+1} = 1. \end{equation}

Finally, plug in the result from the inner function into the outer function to obtain the desired answer: \begin{equation} \lim_{u\to 1} e^{u} = e. \end{equation}

Diatche
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You have correctly written $$ \lim_{x\to 0}(1+\sin x)^{1/x} = \lim_{x\to 0} f(x)^{g(x)} $$ where $f(x)=(1+\sin x)^{1/\sin x}$ and $g(x)=(\sin x)/x$.

Since $f(x) \to e$ and $g(x)\to 1$ and $(a,b)\mapsto a^b$ is known to be continuous at $(e,1)$ you can safely take limits of $f$ and $g$ separately, and thereby find that the original limit is $e$.

hmakholm left over Monica
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Limits of $f(x)^{g(x)}$ in an indeterminate form are usually best treated by computing the limit of $\log(f(x)^{g(x)})=g(x)\log f(x)$. If $l$ is the limit you find, then the original one is $e^l$, when $l$ is finite; if $l=-\infty$, the original limit is $0$; if $l=\infty$, the original limit is $\infty$. This follows from the continuity of the exponential function and of its inverse, the (natural) logarithm.

In your case, you have to find $$ \lim_{x\to0}\log\bigl((1+\sin x)^{1/x}\bigr)= \lim_{x\to0}\frac{\log(1+\sin x)}{x}= \lim_{x\to0}\frac{\sin x+o(\sin x)}{x}= \lim_{x\to0}\frac{x+o(x)}{x}=1 $$ (using $\log(1+t)=t+o(t)$ and $\sin x=x+o(x)$). Thus $$ \lim_{x\to0}(1+\sin x)^{1/x}=e^1=e $$

egreg
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