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

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 , and so I can differentiate the top and the bottom with respect to :

.

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 , then I’ve just shown that (assuming that the limit exists in the first place, of course). That means that or . Well, clearly the limit of this nonnegative function can’t be negative, and so we conclude that the limit is equal to .

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*Related*

*Posted by John Quintanilla on September 24, 2015*

https://meangreenmath.com/2015/09/24/different-ways-of-computing-a-limit-part-2/

## Joseph Nebus

/ September 24, 2015Ooh, yes, that’s a very nice argument.

## Maya Quinn

/ September 26, 2015In 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.