# How do you find #lim (sqrt(y+1)+sqrt(y-1))/y# as #y->oo# using l'Hospital's Rule?

To find (\lim_{y \to \infty} \frac{\sqrt{y+1} + \sqrt{y-1}}{y}) using L'Hôpital's Rule, follow these steps:

- Recognize that (\frac{\sqrt{y+1} + \sqrt{y-1}}{y}) is of an indeterminate form ((\frac{\infty}{\infty})) as (y) approaches infinity.
- Apply L'Hôpital's Rule, which states that if the limit of the quotient of two functions is in an indeterminate form ((\frac{0}{0}) or (\frac{\infty}{\infty})), then the limit of their derivatives will be the same as the original limit.
- Differentiate both the numerator and the denominator separately.
- Evaluate the limit of the resulting expression as (y) approaches infinity.

Let's differentiate:

[ \lim_{y \to \infty} \frac{\sqrt{y+1} + \sqrt{y-1}}{y} = \lim_{y \to \infty} \frac{\frac{d}{dy}(\sqrt{y+1}) + \frac{d}{dy}(\sqrt{y-1})}{\frac{d}{dy}(y)} ]

[ = \lim_{y \to \infty} \frac{\frac{1}{2\sqrt{y+1}} + \frac{1}{2\sqrt{y-1}}}{1} ]

Now, as (y) approaches infinity, both (\sqrt{y+1}) and (\sqrt{y-1}) approach infinity, so their derivatives approach zero:

[ = \frac{0 + 0}{1} = 0 ]

Therefore, (\lim_{y \to \infty} \frac{\sqrt{y+1} + \sqrt{y-1}}{y} = 0).

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