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How can I evaluate the following limit

$$ \lim_{x \rightarrow \infty} \frac{\sqrt{4 + x^2} + x}{x^2 \sin \left(\frac{1}{x}\right)}? $$


When I divided by $x^2$ the numerator and the denominator it gives me $\infty$. Am I right?

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  • $\begingroup$ Note that: $$0\leqslant|x|+x\lt\sqrt{4+x^2}+x\lt\sqrt{4+4|x|+x^2}+x=2+|x|+x\leqslant2(|x|+1)$$ $\endgroup$ – Invisible May 30 at 12:16
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divide both the numerator and the denominator by $x$ and you get: $$\lim_{x \rightarrow \infty} \frac{\sqrt{4 + x^2} + x}{x^2 \sin (\frac{1}{x})}= \lim_{x \rightarrow \infty} \frac{\sqrt{\frac4{x^2} + 1} + 1}{x \sin (\frac{1}{x})} $$ $\lim_{x\to\infty}x\sin\left(\frac1x\right)=1$ is very known limit and it is obvious that $\lim_{x\to\infty}\sqrt{\frac4{x^2} + 1} + 1=\sqrt{1} + 1=2$

hence you have \begin{equation*} \lim_{x \rightarrow \infty} \frac{\sqrt{4 + x^2} + x}{x^2 \sin (\frac{1}{x})} =\frac{2}{1}=2\end{equation*}


the reason you are wrong is because when you divide by $x^2$ you get $$\lim_{x \rightarrow \infty} \frac{\sqrt{4 + x^2} + x}{x^2 \sin (\frac{1}{x})}=\lim_{x \rightarrow \infty} \frac{\sqrt{\frac4{x^4} + \frac1{x^2}} + \frac1x}{\sin (\frac{1}{x})}=\frac{0}{0}$$this is an indeterminate form, and not equal $\infty$

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We have $\lim_{x\rightarrow\infty}\dfrac{1}{x\sin(1/x)}=1$ and $\lim_{x\rightarrow\infty}\dfrac{\sqrt{4+x^{2}}+x}{x}=\lim_{x\rightarrow\infty}\sqrt{\dfrac{4}{x^{2}}+1}+1=2$.

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There are a few things you need to take into account when finding this limit. First of all, recall that $\frac{1}{x}=\frac{\frac{1}{x}}{1}$. Secondly, if $\theta=\frac{1}{x}$ and $x\rightarrow \infty$, then $\theta\rightarrow 0$. And thirdly, we're going to use this simple fact $\lim_\limits{\theta \rightarrow 0}\frac{\theta}{\sin\theta}=1$. Here's the entire solution:

\begin{align} \lim_{x \rightarrow \infty} \frac{\sqrt{4 + x^2} + x}{x^2 \sin \frac{1}{x}} &=\lim_{x \rightarrow \infty} \left(\frac{\sqrt{4 + x^2} + x}{x}\cdot \frac{1}{x\sin \frac{1}{x}}\right)\\ &=\lim_{x \rightarrow \infty} \left[\left(\frac{\sqrt{4 + x^2}}{x}+\frac{x}{x}\right)\cdot \frac{\frac{1}{x}}{\sin \frac{1}{x}}\right]\\ &=\lim_{x \rightarrow \infty} \left(\frac{\sqrt{4 + x^2}}{\sqrt{x^2}}+1\right)\cdot \lim_{x \rightarrow \infty}\frac{\frac{1}{x}}{\sin \frac{1}{x}}\\ &=\lim_{x \rightarrow \infty} \left(\sqrt{\frac{4}{x^2} + 1}+1\right)\cdot \lim_{\theta \rightarrow 0}\frac{\theta}{\sin \theta}\\ &=\left(\sqrt{0 + 1}+1\right)\cdot 1\\ &=\left(1+1\right)\cdot 1\\ &=2 \end{align}


Now, let's actually prove $\lim_\limits{\theta \rightarrow 0}\frac{\theta}{\sin\theta}=1$ because I don't like to leave things unjustified. And I sure hope that you remember one of the most fundamental trigonometric limits in elementary calculus $\lim_\limits{\theta \rightarrow 0}\frac{\sin\theta}{\theta}=1$ (proof is here):

$$ \lim_\limits{\theta \rightarrow 0}\frac{\theta}{\sin\theta}= \lim_\limits{\theta \rightarrow 0}\frac{1}{\frac{\sin\theta}{\theta}}= \frac{\lim_\limits{\theta \rightarrow 0} 1}{\lim_\limits{\theta \rightarrow 0} \frac{\sin\theta}{\theta}}=\frac{1}{1}=1 $$

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