Ytdyklly

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.

$$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}$$

You'll notice that using L'Hopital's rule flips the value of the top to the bottom. For example, using it once returns:

$$limlimits_{xto infty} frac{sqrt{x^2 + 1}}{x}$$

And doing it again returns you to the beginning:

$$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}$$

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?

Hint: Divide the numerator and denominator by $x $ and apply the limit.

$$frac{x}{sqrt{x^2 + 1}}=frac{1}{sqrt{1 + frac{1}{x^2}}}$$

Hint

Simply use $${xover x+1}={xover sqrt{x^2+2x+1}}<{xover sqrt{x^2+1}}<1$$for large enough $x>0$.

By your own reasoning, you have the following:
$$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}=limlimits_{xto infty} frac{sqrt{x^2 + 1}}{x}$$

Now, the left side is clearly the reciprocal of the right side, so we have:
$$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}=frac{1}{limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}}$$

(Note that doing this manipulation assumes that $limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}$ 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 $xinBbb{R}$. Thus, because this is a bounded, always increasing function, the limit as $xto infty$ must converge to some real number, so our assumption in this manipulation is valid.)

Cross-multiply:
$$left(limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}right)^2=1$$

Take the square root:
$$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}=pm 1$$

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:
$$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}=1$$

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.

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 $ttoinfty$ so $$lim_{xtoinfty} frac{x}{sqrt{x^2+1}} = lim_{ttoinfty} tanh t = lim_{ttoinfty}frac{e^t - e^{-t}}{e^t+e^{-t}} = lim_{ttoinfty}frac{e^{2t}-1}{e^{2t}+1} = 1$$