Infinite limits using L'Hospital's Rule

[Asphyxiated]

Homework Statement

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

Homework Equations

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?

 

[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?