# Question #fa661

$\textcolor{red}{{\lim}_{x \rightarrow \infty} {x}^{\sin} \left(\frac{1}{x}\right) = 1}$

#### Explanation:

The solution goes like this

From the given ${\lim}_{x \rightarrow \infty} {x}^{\sin} \left(\frac{1}{x}\right)$

We test first. ${\lim}_{x \rightarrow \infty} {x}^{\sin} \left(\frac{1}{x}\right)$ this results to ${\infty}^{0}$

We apply logarithms

Let $y = {x}^{\sin} \left(\frac{1}{x}\right)$

Take the logarithm of both sides

$y = {x}^{\sin} \left(\frac{1}{x}\right)$

$\ln y = \ln {x}^{\sin} \left(\frac{1}{x}\right)$

$\ln y = \sin \left(\frac{1}{x}\right) \cdot \ln x$

$\ln y = \sin \left(\frac{1}{x}\right) \cdot \ln x$

Test the limit of $\ln y$ and it takes the form $0 \cdot \infty$

We arrange it so that it will be $\frac{\infty}{\infty}$

${\lim}_{x \rightarrow \infty} \ln y = {\lim}_{x \rightarrow \infty} \frac{\ln x}{\frac{1}{\sin} \left(\frac{1}{x}\right)} = \frac{\infty}{\infty}$

Use continuous derivatives on both numerator and denominator until we come up with a real number.

${\lim}_{x \rightarrow \infty} \ln y = {\lim}_{x \rightarrow \infty} \frac{\frac{1}{x}}{\frac{\sin \left(\frac{1}{x}\right) \cdot 0 - 1 \cdot \cos \left(\frac{1}{x}\right) \left(- \frac{1}{x} ^ 2\right)}{{\sin}^{2} \left(\frac{1}{x}\right)}}$

${\lim}_{x \rightarrow \infty} \ln y = {\lim}_{x \rightarrow \infty} \frac{\frac{1}{x}}{\frac{1 \cdot \cos \left(\frac{1}{x}\right) \left(\frac{1}{x} ^ 2\right)}{{\sin}^{2} \left(\frac{1}{x}\right)}} = {\lim}_{x \rightarrow \infty} \frac{{\sin}^{2} \left(\frac{1}{x}\right)}{\frac{1}{x} \cdot \cos \left(\frac{1}{x}\right)} = \frac{0}{0 \cdot 1} = \frac{0}{0}$

${\lim}_{x \rightarrow \infty} \ln y = \frac{0}{0}$

Differentiate again

${\lim}_{x \rightarrow \infty} \ln y =$
$= {\lim}_{x \rightarrow \infty} \frac{2 \sin \left(\frac{1}{x}\right) \cos \left(\frac{1}{x}\right) \cdot \left(- \frac{1}{x} ^ 2\right)}{\frac{1}{x} \left(- \sin \left(\frac{1}{x}\right) \left(- \frac{1}{x} ^ 2\right)\right) + \cos \left(\frac{1}{x}\right) \left(- \frac{1}{x} ^ 2\right)}$

$= {\lim}_{x \rightarrow \infty} \frac{2 \sin \left(\frac{1}{x}\right) \cos \left(\frac{1}{x}\right)}{- \frac{1}{x} \cdot \sin \left(\frac{1}{x}\right) + \cos \left(\frac{1}{x}\right)} = \frac{2 \cdot 0 \cdot 1}{0 + 1} = \frac{0}{1} = 0$

We now have

${\lim}_{x \rightarrow \infty} \ln y = 0$

Finally,

${\lim}_{x \rightarrow \infty} y = {e}^{0} = 1$

And therefore

$\textcolor{red}{{\lim}_{x \rightarrow \infty} {x}^{\sin} \left(\frac{1}{x}\right) = 1}$

God bless....I hope the explanation is useful.