Question

find the limit. (hint: try multiplying and dividing by the conjugate.)
\\(\lim_{x\to+\infty}(\sqrt{x + 36}-\sqrt{x + 81})\\)
\\(\lim_{x\to+\infty}(\sqrt{x + 36}-\sqrt{x + 81})=\square\\) (simplify your answer.)

Explanation:

Step1: Multiply and divide by conjugate

Multiply and divide $\lim_{x
ightarrow+\infty}(\sqrt{x + 36}-\sqrt{x + 81})$ by $\sqrt{x + 36}+\sqrt{x + 81}$:
\[

$$\begin{align*} &\lim_{x ightarrow+\infty}\frac{(\sqrt{x + 36}-\sqrt{x + 81})(\sqrt{x + 36}+\sqrt{x + 81})}{\sqrt{x + 36}+\sqrt{x + 81}}\\ =&\lim_{x ightarrow+\infty}\frac{(x + 36)-(x + 81)}{\sqrt{x + 36}+\sqrt{x + 81}} \end{align*}$$

\]

Step2: Simplify the numerator

Simplify the numerator $(x + 36)-(x + 81)$:
\[

$$\begin{align*} (x + 36)-(x + 81)&=x+36 - x-81\\ &=- 45 \end{align*}$$

\]
So the limit becomes $\lim_{x
ightarrow+\infty}\frac{-45}{\sqrt{x + 36}+\sqrt{x + 81}}$.

Step3: Analyze the limit as $x

ightarrow+\infty$
As $x
ightarrow+\infty$, $\sqrt{x + 36}
ightarrow+\infty$ and $\sqrt{x + 81}
ightarrow+\infty$. Then $\sqrt{x + 36}+\sqrt{x + 81}
ightarrow+\infty$.
We know that $\lim_{x
ightarrow+\infty}\frac{-45}{\sqrt{x + 36}+\sqrt{x + 81}} = 0$ since the numerator is a constant $-45$ and the denominator approaches infinity.

Answer:

$0$