# How to find an analytic solution to $\lim_{x \to +\infty} \frac{x+\sin(x^2)}{\sqrt{x^2+1}}$

To solve this limit:

$$\lim_{x \to +\infty} \space \frac{x+\sin(x^2)}{\sqrt{x^2+1}}$$

At the beginning I didn't know how to start. Then I thought, no matter the value that $x$ takes, $sin(x^2)$ will always be between $-1$ and $1$. So for large values of $x$, $sin(x^2)$ is insignificant. One can rewrite:

$$\lim_{x \to +\infty} \space \frac{x}{\sqrt{x^2+1}}$$

And now it's easy to find the limit:

$$\lim_{x \to +\infty} \space \frac{x}{\sqrt{x^2+1}} = \lim_{x \to +\infty} \space \frac{\sqrt{x^2}}{\sqrt{x^2+1}} = \lim_{x \to +\infty} \space \sqrt{\frac{x^2}{x^2+1}} = \lim_{x \to +\infty} \space \sqrt{\frac{x^2}{x^2}}=1$$

But I know that the justification that allowed me to find the limit this way is not an analytic justification. How can I find this limit on an analytic basis?

Thanks

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Your title and problem do not match, is $n$ supposed to be $x$? Regards –  Amzoti Jan 20 '13 at 18:57
Yes!My mistake. Thanks to edit –  João Jan 20 '13 at 18:59
The only justification you need is to say "By the Squeeze Theorem". That is what allows you to bound $-1\leq\sin(x^2)\leq1$. –  Clayton Jan 20 '13 at 18:59

Write, for $x>0$ $${x+\sin(x^2)\over\sqrt{x^2+1}}= {{1\over x}\cdot(x+\sin(x^2))\over{1\over x}\sqrt{x^2+1}}= {{1+{\sin(x^2)\over x} } \over\sqrt{1+{1\over x^2}}}.$$

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You know that $\lim \limits_{x \to +\infty} \left(\dfrac{x+\sin(x^2)}{\sqrt{x^2+1}}\right)= \lim \limits_{x\to +\infty}\left(\dfrac{x}{\sqrt{x^2+1}}+\dfrac{\sin(x^2)}{\sqrt{x^2+1}}\right)$.
Now if both $\lim \limits_{x\to +\infty}\left(\dfrac{x}{\sqrt{x^2+1}}\right)$ and $\lim \limits_{x\to +\infty}\left(\dfrac{\sin(x^2)}{\sqrt{x^2+1}}\right)$ exist, then $$\lim \limits_{x\to +\infty}\left(\dfrac{x}{\sqrt{x^2+1}}+\frac{\sin(x^2)}{\sqrt{x^2+1}}\right)=\lim \limits_{x\to +\infty}\left(\frac{x}{\sqrt{x^2+1}}\right )+ \lim \limits_{x\to +\infty}\left(\frac{\sin(x^2)}{\sqrt{x^2+1}}\right)$$

You've proved that the first one on the RHS exists and equals $1$.

To show that the second one exists and equals $0$ it suffices to prove that $\lim \limits_{x\to +\infty}\left(\left | \dfrac{\sin(x^2)}{\sqrt{x^2+1}} \right |\right)=0$ and that's a consequence of the squeeze theorem because for all $x\in \mathbb{R},$ $$0\leq \left | \dfrac{\sin(x^2)}{\sqrt{x^2+1}} \right |\leq \left | \frac{1}{\sqrt{x^2+1}} \right |$$ and $\lim \limits_{x\to +\infty}\left(\left | \frac{1}{\sqrt{x^2+1}} \right |\right)=0$.

Therefore $$\lim \limits_{x\to +\infty}\left(\dfrac{x}{\sqrt{x^2+1}}+\frac{\sin(x^2)}{\sqrt{x^2+1}}\right)=\lim\limits_{x\to +\infty}\left(\frac{x}{\sqrt{x^2+1}}\right )+ \lim\limits_{x\to +\infty}\left(\frac{\sin(x^2)}{\sqrt{x^2+1}}\right)=1+0.$$

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If $h(x)\leq f(x)\leq g(x) \forall x\in \Bbb R\implies \lim_{x\to\infty}h(x)\leq \lim_{x\to\infty}f(x)\leq \lim_{x\to\infty}g(x)$

Here, $$h(x)=\frac{x-1}{\sqrt{x^2+1}},f(x)=\frac{x-\sin(x^2)}{\sqrt{x^2+1}})$$ and $$g(x)=\frac{x+1}{\sqrt{x^2+1}}$$

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