Is there a way to find this limit algebraically? $\lim\limits_{x\to \infty} \frac{x}{\sqrt{x^2 + 1}}$ I'm a Calculus I student and my teacher has given me a set of problems to solve with L'Hoptial's rule. Most of them have been pretty easy, but this one has me stumped. 
$$\lim\limits_{x\to \infty} \frac{x}{\sqrt{x^2 + 1}}$$ 
You'll notice that using L'Hopital's rule flips the value of the top to the bottom. For example, using it once returns: 
$$\lim\limits_{x\to \infty} \frac{\sqrt{x^2 + 1}}{x}$$ 
And doing it again returns you to the beginning: 
$$\lim\limits_{x\to \infty} \frac{x}{\sqrt{x^2 + 1}}$$ 
I of course plugged it into my calculator to find the limit to evaluate to 1, but I was wondering if there was a better way to do this algebraically?
 A: By your own reasoning, you have the following:
$$\lim\limits_{x\to \infty} \frac{x}{\sqrt{x^2 + 1}}=\lim\limits_{x\to \infty} \frac{\sqrt{x^2 + 1}}{x}$$
Now, the left side is clearly the reciprocal of the right side, so we have:
$$\lim\limits_{x\to \infty} \frac{x}{\sqrt{x^2 + 1}}=\frac{1}{\lim\limits_{x\to \infty} \frac{x}{\sqrt{x^2 + 1}}}$$
(Note that doing this manipulation assumes that $\lim\limits_{x\to \infty} \frac{x}{\sqrt{x^2 + 1}}$ converges to a real number. However, you can use the first derivative to show this is an always increasing function and then use basic algebra to show that $\frac{x}{\sqrt{x^2 + 1}} < 1$ for all $x\in\Bbb{R}$. Thus, because this is a bounded, always increasing function, the limit as $x\to \infty$ must converge to some real number, so our assumption in this manipulation is valid.)
Cross-multiply:
$$\left(\lim\limits_{x\to \infty} \frac{x}{\sqrt{x^2 + 1}}\right)^2=1$$
Take the square root:
$$\lim\limits_{x\to \infty} \frac{x}{\sqrt{x^2 + 1}}=\pm 1$$
However, it is easy to show that $\frac{x}{\sqrt{x^2 + 1}}>0$ for all $x > 0$. Therefore, there's no way the limit can be a negative number like $-1$. Thus, the only possibility we have left is $+1$, so:
$$\lim\limits_{x\to \infty} \frac{x}{\sqrt{x^2 + 1}}=1$$
A: When computing the limit of rational functions, as is the case for $$\lim_{x \rightarrow \infty} \frac{x}{\sqrt{x^2 +1}},$$ you want to divide the top and bottom by the highest degree in the denominator, which in this case is $x$. Since $x \rightarrow +\infty$, so $x$ is always positive (at least, near where we are worried about) I claim that $x = \sqrt{x^2}$. So, if we divide the top and bottom by $x$, we get $$\lim_{x \rightarrow \infty} \frac{x}{\sqrt{x^2 +1}} = \lim_{x \rightarrow \infty} \frac{1}{\sqrt{1 + 1/x^2}}.$$ You should be able to compute the limit from here.
Whenever you see a monomial in the numerator with the square root of a polynomial in the denominator, you should consider this method. Of course, keep in mind that you'll have to tweak it slightly if $x \rightarrow -\infty$! Try to see if you can figure out what would change in that case.
A: Hint: Divide the numerator and denominator by  $x $ and apply the limit.
$$\frac{x}{\sqrt{x^2 + 1}}=\frac{1}{\sqrt{1 + \frac{1}{x^2}}}$$
A: Hint
Simply use $${x\over x+1}={x\over \sqrt{x^2+2x+1}}<{x\over \sqrt{x^2+1}}<1$$for large enough $x>0$.
A: Set $x = \sinh t$. We have
$$\frac{x}{\sqrt{x^2+1}}= \frac{\sinh t}{\sqrt{1+\sinh^2t}} = \frac{\sinh t}{\cosh t} = \tanh t$$
$x \to \infty$ is equivalent to $t\to\infty$ so $$\lim_{x\to\infty} \frac{x}{\sqrt{x^2+1}} = \lim_{t\to\infty} \tanh t = \lim_{t\to\infty}\frac{e^t - e^{-t}}{e^t+e^{-t}} = \lim_{t\to\infty}\frac{e^{2t}-1}{e^{2t}+1} = 1$$
