Proving $ \frac{x^2\sin x}{3\sqrt{x+1}}$ DNE

Question

Not sure how to prove this; WolframAlpha tells me that the result is indeterminate. My attempt: \begin{align*} \lim_{x\to\infty} \frac{x^2\sin x}{3\sqrt{x+1}}&=\lim_{x\to\infty}\frac{x^2\sin x}{3\sqrt{x+1}}\cdot \frac{1/x^2}{1/x^2} \\&=\lim_{x\to\infty}\frac{1\cdot \sin x}{3\sqrt{x+1}\cdot \frac{1}{x^2}} \\&=\lim_{x\to\infty}\frac{\sin x}{3\sqrt{x+1}\cdot \frac{1}{\sqrt{x^4}}} \\&=\lim_{x\to\infty}\frac{\sin x}{3\sqrt{\frac{x+1}{x^4}}} \\&=\lim_{x\to\infty}\frac{\sin x}{3\sqrt{1/x^3+1/x^4}} \end{align*} Any hints using introductory calculus principles (no L'Hopital's Rule or further) would be appreciated.

Answer 1

$\sin(x)$ oscillates between $+1$ and $-1$, while $x^2/(3 \sqrt{x+1}) > x$ when $x$ is large enough. So for any $L$ there are arbitrarily large $x$ where your function is far away from $L$, therefore the limit does not exist. The limit is also not $+\infty$, because there are arbitrarily large $x$ where the value is negative, and not $-\infty$, because there are arbitrarily large $x$ where the value is positive.

Answer 2

It becomes really easy to see this by graphs. To get an approximate graph of $\sin(x)\cdot f(x)$ , we draw $f(x)$ and $-f(x)$. Then we draw a smooth zigzag curve between the two curves to get the approximate graph.

I am presenting a few of them:

$x\sin(x)$

$\frac{x^2}{100}\sin(x)$

$\ln(x)\sin(x)$

Now, the graph of $f(x)=\frac{x^{2}}{3\sqrt{x+1}}$ tends to infinity at $x\to\infty$ and $-f(x)$ tends to negative infinity as $x\to\infty$

As the graph of $f(x)\sin(x)$ oscillates, so it would not have any well defined limit as $x\to\infty$

That is an intuitive answer for you. Cheers :)