# Infinite limits using L'Hospital's Rule

1. May 19, 2010

### Asphyxiated

1. The problem statement, all variables and given/known data

$$\lim_{y \to \infty} \frac {\sqrt{y+1}+\sqrt{y-1}}{y}$$

2. Relevant equations

3. The attempt at a solution

By L'Hospital's Rule:

$$\lim_{y \to \infty} \frac {\sqrt{y+1}+\sqrt{y-1}}{y} = \lim_{y \to \infty} \frac {\frac{1}{2}(y+1)^{-1/2}+\frac{1}{2}(y-1)^{-1/2}}{1}$$

which is just this:

$$\lim_{y \to \infty} \frac{1}{2}(y+1)^{-1/2}+\frac{1}{2}(y-1)^{-1/2}$$

and the answer is suppose to be 0, is this because you are basically taking the reciprocal of infinity which is infinitely or arbitrarily close to 0?

2. May 19, 2010

### physicsman2

when you apply the limit, you should get 1/infinity + 1/infinity, which is 0 + 0 = 0.

A small number over a much larger number goes to 0.

Edit: Instead of using L'Hopital's Rule, you could just say that the power of y in the denominator is larger than the power of y in the numerator, so since you're taking the limit as y approaches infinity, you could say the limit is zero. Do you see why?