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 $2$

  • $\begingroup$ $(2x^2+2)\sin(\frac{1}{x^2+1})=\frac{2+2/x^2}{1+1/x^2}\frac{\sin(1/(x^2+1))}{1/(x^2+1)}$. The second factor tends to 1, the first to 2. $\endgroup$ – Hellen Oct 1 '17 at 13:02
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    $\begingroup$ Do you mean sine or is sen some other function? $\endgroup$ – MrYouMath Oct 1 '17 at 13:59
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    $\begingroup$ @MrYouMath In my country (Italy) it's common to use “sen” for the sine function, because it's called seno. I usually write “sin”, considering it the abbreviation of the Latin sinus. $\endgroup$ – egreg Oct 1 '17 at 14:07
  • $\begingroup$ @egreg: Thank you for this interesting fact :). $\endgroup$ – MrYouMath Oct 1 '17 at 14:27

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

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

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


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:


Therefore for $x$ very large


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.


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