Question

find the limit. (if the limit is infinite, enter ∞ or -∞, as appropriate. if the limit does
lim _{x
ightarrow infty} \frac{36-sqrt{x}}{6+sqrt{x}}

Explanation:

Step1: Divide numerator and denominator by $\sqrt{x}$

$$\lim_{x ightarrow\infty}\frac{\frac{36}{\sqrt{x}} - 1}{\frac{6}{\sqrt{x}}+1}$$

Step2: Use the limit property $\lim_{x

ightarrow\infty}\frac{c}{x^n}=0$ ($n>0,c$ is a constant)
As $x
ightarrow\infty$, $\lim_{x
ightarrow\infty}\frac{36}{\sqrt{x}} = 0$ and $\lim_{x
ightarrow\infty}\frac{6}{\sqrt{x}}=0$.
So the limit becomes $\frac{0 - 1}{0 + 1}$

Answer:

$-1$