# Different ways of computing a limit (Part 2)

One of my colleagues placed the following problem on an exam for his Calculus II course…

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

and was impressed by the variety of correct responses that he received. I thought it would be fun to discuss some of the different ways that this limit can be computed.

Method #2. Using L’Hopital’s Rule. The limit has the indeterminant form $\infty/\infty$, and so I can differentiate the top and the bottom with respect to $x$:

$\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x} = \displaystyle \lim_{x \to \infty} \frac{ \displaystyle \frac{d}{dx} \left( \sqrt{x^2+1} \right) }{\displaystyle \frac{d}{dx} \left( x \right)}$

$= \displaystyle \lim_{x \to \infty} \frac{ \displaystyle \frac{1}{2} \left( x^2+1 \right)^{-1/2} \cdot 2x }{1}$

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

Oops… it looks like I just got the reciprocal of the original limit! Indeed, if I use L’Hopital’s Rule again, I’ll just return back to the original limit.

So that doesn’t look very helpful… except it is. If I define the value of this limit to be equal to $L$, then I’ve just shown that $L = 1/L$ (assuming that the limit exists in the first place, of course). That means that $L = 1$ or $L = -1$. Well, clearly the limit of this nonnegative function can’t be negative, and so we conclude that the limit is equal to $1$.

## 2 thoughts on “Different ways of computing a limit (Part 2)”

1. In a similar vein: If you are supposing the limit converges to L in the first place, then its square will converge to L^2.

But squaring returns \lim_{x to \infty} of 1 + 1/x^2 = 1.

So L^2 = 1 etc.

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