# What is the limit of x^2sin(1/x) as x approaches infinity?

Jan 4, 2018

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

#### Explanation:

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

If we use direct substitution we'd get the indeterminate form $\infty \cdot 0$

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

We can use L'Hospital only if we have the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$. We can however rewrite the limit to accommodate for that.

If we rewrite the limit as...

${\lim}_{x \to \infty} \sin \frac{\frac{1}{x}}{\frac{1}{x} ^ 2}$

And use direct substitution we'd get the indeterminate form of $\frac{0}{0}$

Now we apply L'Hospital's Rule

${\lim}_{x \to \infty} \sin \frac{\frac{1}{x}}{\frac{1}{x} ^ 2} = {\lim}_{x \to \infty} \frac{- \cos \frac{\frac{1}{x}}{x} ^ 2}{- \frac{2}{x} ^ 3}$

Simplifying:

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

$= \frac{1}{2} \cdot {\lim}_{x \to \infty} \left(x \cos \left(\frac{1}{x}\right)\right)$

Now using direct substitution we'll find the limit to be...

$= \frac{1}{2} \cdot \infty \cdot \cos \left(\frac{1}{\infty}\right) = \frac{1}{2} \cdot \infty \cdot \cos \left(0\right) = \frac{1}{2} \cdot \infty \cdot 1 = \infty$

Jan 5, 2018

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

#### Explanation:

Use:

${\lim}_{t \to 0} \sin \frac{t}{t} = 1$

Letting $t = \frac{1}{x}$ we have:

${\lim}_{x \to \infty} {x}^{2} \sin \left(\frac{1}{x}\right) = {\lim}_{t \to {0}^{+}} \frac{1}{t} ^ 2 \sin \left(t\right)$

$\textcolor{w h i t e}{{\lim}_{x \to \infty} {x}^{2} \sin \left(\frac{1}{x}\right)} = {\lim}_{t \to {0}^{+}} \frac{1}{t} ^ 2 \sin \left(t\right)$

$\textcolor{w h i t e}{{\lim}_{x \to \infty} {x}^{2} \sin \left(\frac{1}{x}\right)} = \left({\lim}_{t \to {0}^{+}} \frac{1}{t}\right) \cdot \left({\lim}_{t \to {0}^{+}} \sin \frac{t}{t}\right)$

$\textcolor{w h i t e}{{\lim}_{x \to \infty} {x}^{2} \sin \left(\frac{1}{x}\right)} = {\lim}_{t \to {0}^{+}} \frac{1}{t} \cdot 1$

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