$\lim_{x\to\infty} (2x^2+2)\operatorname{sin}\left(\frac{1}{x^2+1}\right)$ without using L’Hôpital.

Question

Can I make the following limit without applying L'Hôpital?

$$\lim_{x\to\infty} (2x^2+2)\operatorname{sin}\left(\frac{1}{x^2+1}\right)$$

With L'Hôpital it gives me as a result $2$

Answer 1

You can use the following limit: $$\lim_{t\to 0} \dfrac{\sin(t)}{t} = 1$$.

Here let $t = \dfrac{1}{x^2+1}$. As $x \to \infty$, $t \to 0$.

$$\lim_{x\to \infty}\, (2x^2 + 2) \sin\left(\dfrac{1}{1+x^2}\right) \\= \lim_{t\to 0}\, 2\dfrac{\sin(t)}{t} = 2$$

Answer 2

One can do this types of limits without using L'Hopital without relying on the fact that the first polynomial $(2x^2+2)$ is exactly twice the polynomial in the denominator inside the sine. The only thing that matters to get a limit of $2$ is that the leading term in the first polynomial is twice the leading term of the other. Consider the more general case:

$$ \lim_{x\to\infty}(2x^2+ax+b)\sin\left(\frac{1}{x^2+cx+d}\right)$$

where $a,b,c,d$ are arbitrary constants. Here is how I would do it. For $x$ large enough the argument inside the $\sin()$ becomes very small, so we can approximate it by:

$$\sin\bigg(\frac{1}{x^2+cx+d}\bigg)\approx\frac{1}{x^2+cx+d}$$

Therefore for $x$ very large

$$(2x^2+ax+b)\sin\bigg(\frac{1}{x^2+cx+d}\bigg)\approx(2x^2+ax+b)\frac{1}{x^2+cx+d}\to2$$

One can easily see that the neglected terms in the expansion of the sine go to zero as $O(x^{-6})$, way faster than the polynomial $2x^2+ax+b$ goes to $\infty$, so they do not contribute to the limit.