Let $f(x)=\dfrac{4\sin(x)}{x^2+1}$. How would I find the $\lim_\limits{x\to\infty}\bigg(\dfrac{4\sin(x)}{x^2+1}\bigg)$? I know that $\sin(x)$ oscillates periodically between $1$, and $-1$, so would the limit not exist?
Horizontal Asymptote- Limits at Infinity
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limits
5 Answers
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$\left|\frac{4\sin(x)}{x^2+1}\right|\le\frac{4}{x^2+1}\rightarrow 0$. So your asymptote is $y=0$
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It's true that the limit of sin(x) doesn't exist, but look at the denominator. What happens to the overall expression as x gets large? The limit does exist.
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The oscillations don't matter here since the denominator grows indefinitely. Just note that $$ 0\leq\left|\frac{4\sin(x)}{x^2+1}\right|\leq\frac{4}{x^2+1} $$ and now use the Squeeze Theorem to conclude that the limit as $x\to\infty$ of your function is $0$.
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This might give you more intuition about why the oscillation does not really matter:
On the other hand, mathematically, $$ \left|\frac{4\sin x}{x^2+1}\right|\leq\left|\frac{4}{x^2+1}\right|. $$
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We must use something called the squeeze theorem here. First of all, we know that $$-1 \leq \sin(x) \leq 1$$ $$-4 \leq 4\sin(x) \leq 4$$ $$\frac{-4}{x^2+1}\leq \frac{4\sin(x)}{x^2+1} \leq \dfrac{4}{x^2+1}$$
What the squeeze theorem allows is that if the $\lim_\limits{x\to\infty}\bigg(\dfrac{-4}{x^2+1}\bigg) = \lim_\limits{x\to\infty}\bigg(\dfrac{4}{x^2+1}\bigg)$, then that is also equal to $\lim_\limits{x\to\infty}\bigg(\dfrac{4\sin(x)}{x^2+1}\bigg)$.
$$\lim_\limits{x\to\infty}\bigg(\dfrac{-4}{x^2+1}\bigg) = 0$$ $$\lim_\limits{x\to\infty}\bigg(\frac{4}{x^2+1}\bigg) = 0$$
THUS,
$$\lim_\limits{x\to\infty}\bigg(\dfrac{4\sin(x)}{x^2+1}\bigg) = 0$$