# What is lim_(xrarroo) (e^(2x)sin(1/x))/x^2 ?

Apr 8, 2016

${\lim}_{x \to \infty} \frac{{e}^{2 x} \sin \left(\frac{1}{x}\right)}{x} ^ 2 = \infty$

#### Explanation:

Let $y = \frac{{e}^{2 x} \sin \left(\frac{1}{x}\right)}{x} ^ 2$

$\ln y = \ln \left(\frac{{e}^{2 x} \sin \left(\frac{1}{x}\right)}{x} ^ 2\right)$

$\ln y = \ln {e}^{2 x} + \ln \left(\sin \left(\frac{1}{x}\right)\right) - \ln {x}^{2}$

$\ln y = 2 x \ln e + \ln \left(\sin \left(\frac{1}{x}\right)\right) - 2 \ln x$

$\ln y = 2 x + \ln \left(\sin \left(\frac{1}{x}\right)\right) - 2 \ln x$

${\lim}_{x \to \infty} \left[\ln y = 2 x + \ln \left(\sin \left(\frac{1}{x}\right)\right) - 2 \ln x\right]$

${\lim}_{x \to \infty} \ln y = {\lim}_{x \to \infty} \left[2 x + \ln \left(\sin \left(\frac{1}{x}\right)\right) - 2 \ln x\right]$

${\lim}_{x \to \infty} \ln y = \infty$

${e}^{\ln} y = {e}^{\infty}$

$y = \infty$

Apr 8, 2016

${\lim}_{x \rightarrow \infty} \frac{{e}^{2 x} \sin \left(\frac{1}{x}\right)}{x} ^ 2 = \infty$. Please see the explanation section below.

#### Explanation:

${\lim}_{x \rightarrow \infty} \frac{{e}^{2 x} \sin \left(\frac{1}{x}\right)}{x} ^ 2$

Note that: $\frac{{e}^{2 x} \sin \left(\frac{1}{x}\right)}{x} ^ 2 = {e}^{2 x} / {x}^{3} \cdot \sin \frac{\frac{1}{x}}{\frac{1}{x}}$

Now, as $x \rightarrow \infty$, the first ratio increases without bound, while the second goes to $1$.

${\lim}_{x \rightarrow \infty} \frac{{e}^{2 x} \sin \left(\frac{1}{x}\right)}{x} ^ 2 = {\lim}_{x \rightarrow \infty} {e}^{2 x} / {x}^{3} \cdot {\lim}_{x \rightarrow \infty} \sin \frac{\frac{1}{x}}{\frac{1}{x}}$

$= \infty$

Further Explanation

Here is the reasoning that led to the solution above.

${\lim}_{x \rightarrow \infty} \frac{{e}^{2 x} \sin \left(\frac{1}{x}\right)}{x} ^ 2$ has initial form $\frac{\infty \cdot 0}{\infty}$.

This is an indeterminate form, but we cannot apply l'Hospital's Rule to this form.

We could rewrite it as $\frac{{e}^{2 x}}{{x}^{2} / \sin \left(\frac{1}{x}\right)}$ to get the form $\frac{\infty}{\infty}$ to which we could apply l'Hospital. However, I don't particularly want to take the derivative of that denominator.

Recall that ${\lim}_{\theta \rightarrow 0} \sin \frac{\theta}{\theta} = 1$.

So that ${\lim}_{x \rightarrow \infty} \sin \frac{\frac{1}{x}}{\frac{1}{x}} = 1$.

This is what motivates the rewriting used above.

$\frac{{e}^{2 x} \sin \left(\frac{1}{x}\right)}{x} ^ 2 = {e}^{2 x} / {x}^{3} \cdot \sin \frac{\frac{1}{x}}{\frac{1}{x}}$.

As $x$ increases without bound, ${e}^{x}$ goes to infinity much faster that ${x}^{3}$ (faster than any power of $x$).
So, ${e}^{2 x} = {\left({e}^{x}\right)}^{2}$ blows up even faster.

If you do not have this fact available, use l'Hospital's rule to get

${\lim}_{x \rightarrow \infty} {e}^{2 x} / {x}^{3} = {\lim}_{x \rightarrow \infty} \frac{2 {e}^{2 x}}{3 {x}^{2}}$

$= {\lim}_{x \rightarrow \infty} \frac{8 {e}^{2 x}}{6} = \infty$