# How do you evaluate lim_(xtooo) (5 - x^(1/2))/(5 + x^(1/2)) ?

Jun 12, 2017

#### Explanation:

Given: ${\lim}_{x \to \infty} \frac{5 - {x}^{\frac{1}{2}}}{5 + {x}^{\frac{1}{2}}}$

Because the expression evaluated at the limit yields the indeterminate form $\frac{\infty}{\infty}$, the use of L'Hôpital's rule is warranted.

You apply the rule by differentiating the number and the denominator:

${\lim}_{x \to \infty} \frac{\frac{d \left(\left(5 - {x}^{\frac{1}{2}}\right)\right)}{\mathrm{dx}}}{\frac{d \left(\left(5 + {x}^{\frac{1}{2}}\right)\right)}{\mathrm{dx}}} =$

${\lim}_{x \to \infty} \frac{- \frac{1}{2} {x}^{- \frac{1}{2}}}{\frac{1}{2} {x}^{- \frac{1}{2}}} =$

${\lim}_{x \to \infty} - 1 = - 1$

According to the rule, so goes the original limit:

${\lim}_{x \to \infty} \frac{5 - {x}^{\frac{1}{2}}}{5 + {x}^{\frac{1}{2}}} = - 1$

Jun 12, 2017

I got: $- 1$

#### Explanation:

If you try directly you get:

${\lim}_{x \to \infty} \frac{5 - {x}^{\frac{1}{2}}}{5 + {x}^{\frac{1}{2}}} = - \frac{\infty}{\infty}$

We can use de L'Hospital Rule (deriving top and bottom to get):

${\lim}_{x \to \infty} \frac{5 - {x}^{\frac{1}{2}}}{5 + {x}^{\frac{1}{2}}} = {\lim}_{x \to \infty} \frac{- \frac{1}{2} \cancel{{x}^{- \frac{1}{2}}}}{\frac{1}{2} \cancel{{x}^{- \frac{1}{2}}}} = - \frac{\frac{1}{2}}{\frac{1}{2}} = - 1$