Question

i need help with this problem. it’s calc 1 in high school

Answer 1

Answer:

-1

Explanation:

Given the limit below:

$\lim _{x\to\infty}\mleft(\frac{9-\sqrt{x}}{3+\sqrt{x}}\mright)$

Follow the steps below to find the limit.

Step 1: Divide both the numerator and denominator by √x.

$\lim _{x\to\infty}\mleft(\frac{\frac{9-\sqrt[]{x}}{\sqrt[]{x}}}{\frac{3+\sqrt[]{x}}{\sqrt[]{x}}}\mright)=\lim _{x\to\infty}\mleft(\frac{\frac{9}{\sqrt[]{x}}-1}{\frac{3}{\sqrt[]{x}}+1}\mright)$

Step 2: Next, apply the given rule of limits below:

$\begin{gathered} \lim _{x\to\infty}\mleft(\frac{f(x)}{g(x)}\mright)=\frac{\lim_{x\to\infty}\mleft(f(x)\mright)}{\lim_{x\to\infty}\mleft(g(x)\mright)},g(x)\neq0 \\ \implies\lim _{x\to\infty}\mleft(\frac{\frac{9}{\sqrt[]{x}}-1}{\frac{3}{\sqrt[]{x}}+1}\mright)=\frac{\lim_{x\to\infty}\mleft(\frac{9}{\sqrt[]{x}}-1\mright)}{\lim_{x\to\infty}\mleft(\frac{3}{\sqrt[]{x}}+1\mright)} \end{gathered}$

Step 3: Find the limits.

$\begin{gathered} \lim _{x\to\infty}(\frac{9}{\sqrt[]{x}}-1)=-1 \\ \lim _{x\to\infty}(\frac{3}{\sqrt[]{x}}+1)=1 \end{gathered}$

Therefore:

$\lim _{x\to\infty}(\frac{9-\sqrt[]{x}}{3+\sqrt[]{x}})=-\frac{1}{1}=-1$

The limit as x tends to infinity is -1.