What is the limit of

Solution 1:

$$\left(1+\sin\frac{1}{x}\right)^x=e^{x\ln\left(1+\sin \frac{1}{x}\right)}.$$

You have that for $x\to \infty $, $\sin \frac{1}{x}\sim \frac{1}{x}$, therefore

$$\lim_{x\to \infty }x\ln\left(1+\sin \frac{1}{x}\right)=\lim_{x\to \infty }\frac{\ln\left(1+ \frac{1}{x}\right)}{\frac{1}{x}}=\lim_{u\to 0}\frac{\ln(1+u)}{u}=1.$$

Moreover $x\mapsto e^x$ is continuous at $x=1$, and thus

$$\lim_{x\to\infty }e^{x\ln\left(1+\sin \frac{1}{x}\right)}=e^1.$$

Solution 2:

Take a logarithm. Let $A$ be the limit. Then

\begin{align} A &= \lim_{x\to\infty} \left(1+\sin\frac{1}{x}\right)^x \\\\ &=\lim_{x\to\infty} \,\,\exp\left(x\ln\left(1+\sin\frac{1}{x}\right)\right)\\\\ &=\lim_{y\to0} \,\,\exp\left(\frac{\ln\left(1+\sin y\right)}{y}\right)\\\\ \end{align}

Using the Taylor series for $\ln (1+z)$ with $z=\sin y$ we find that \begin{align} A&=\,\,\lim_{y\to 0}\,\,\exp\left(\displaystyle\frac{\sin y - \frac{\sin^2 y}{2}+\frac{\sin^3 y}{3} -\frac{\sin^4 y}{4}+\cdots}{y}\right)\\ \end{align}

Remember that $$\lim_{y\to 0} \frac{\sin y}{y}=1 \qquad\,\, \text{and, }\forall \, n >1, \,n \in \mathbb{N}\text{,} \qquad \lim_{y\to 0} \frac{\sin^n y}{ny}=0$$

So in the exponent we have $1$ plus an alternating sum of terms that go to zero, which gives

$$A=\lim_{y\to 0} \,\,\exp(1+0)=\boxed{e}$$

Solution 3:

$$\left(1+\sin\frac{1}{x}\right)^x = \left(1+\frac{1}{x} + \mathcal{O}\left(\frac{1}{x^3}\right)\right)^x,$$ as $x \rightarrow \infty$. Thus, $$ \lim \limits_{x \rightarrow \infty} \left(1+\sin\frac{1}{x}\right)^x = \lim \limits_{x \rightarrow \infty} \left(1+\frac{1}{x}\right)^x = e.$$