# Thread: lhospital

1. ## lhospital

lim x->oo (x^2)sin(1/x)

2. Originally Posted by jeph
lim x->oo (x^2)sin(1/x)
Do not use it!

$-1 \leq \sin \frac{1}{x} \leq 1$ for all $x\not = 0$

Thus,

$-x^2 \leq x^2 \sin \frac{1}{x} \leq x^2$

Now, $\lim_{x\to 0} -x^2 = \lim_{x\to 0}x^2 =0$

By squeeze theorem,
$\lim_{x\to 0} x^2 \sin \frac{1}{x} =0$

3. are you sure? o.o

the answer my prof provided was oo

4. Originally Posted by jeph
are you sure? o.o

the answer my prof provided was oo
I did it for $x\to 0$ instead of $x\to \infty$.

In that case,
$\lim_{x\to \infty} x^2\sin \frac{1}{x} = +\infty$

But we do not need to use that rule.

Instead for sufficiently large $x$ consider the reciprocal, i.e. $\frac{1}{x^2\sin \frac{1}{x}}$

We can think of this reciprocal function as a composition $f\circ g$ where $g = \frac{1}{x}$ and $f=\frac{x^2}{\sin x}$. But at $x=0$ we define $f$ to be zero. Why? Because $\lim_{x\to 0}f = \lim_{x\to 0} \frac{x^2}{\sin x} = \lim_{x\to 0} x \cdot \frac{x}{\sin x} = 0 \cdot 1 =0$. This makes $f$ continous at zero.

Now $\lim_{x\to \infty} g = 0$ and $f$ is continous at 0. By the limit composition rule the limit $\lim_{x\to 0}f\circ g = f(0) = 0$.

We have shown that $\lim_{x\to +\infty} \frac{1}{x^2\sin \frac{1}{x}} =0$. Which implies that its reciprocal is $+\infty$ that is, $\lim_{x\to +\infty} x^2 \cdot \sin \frac{1}{x} = \infty$.