Question

homework2: problem 12 (1 point) evaluate the limit: $lim_{x
ightarrow16}left(\frac{1}{sqrt{x}-4}-\frac{8}{x - 16}
ight)=square$ preview my answers submit answers you have attempted this problem 0 times. you have unlimited attempts remaining.

Explanation:

Step1: Find a common denominator

First, rewrite the two - fraction expression with a common denominator. The common denominator of $\sqrt{x}-4$ and $x - 16$ is $(\sqrt{x}-4)(x - 16)$. We know that $x-16=(\sqrt{x}+4)(\sqrt{x}-4)$. So, $\frac{1}{\sqrt{x}-4}-\frac{8}{x - 16}=\frac{x - 16-8(\sqrt{x}-4)}{(\sqrt{x}-4)(x - 16)}=\frac{x-8\sqrt{x}+16}{(\sqrt{x}-4)(x - 16)}$.

Step2: Factor the numerator

Factor the numerator $x - 8\sqrt{x}+16$. Let $t=\sqrt{x}$, then $x - 8\sqrt{x}+16=t^{2}-8t + 16=(t - 4)^{2}=(\sqrt{x}-4)^{2}$. So the expression becomes $\frac{(\sqrt{x}-4)^{2}}{(\sqrt{x}-4)(x - 16)}$.

Step3: Simplify the expression

Cancel out the common factor $\sqrt{x}-4$ (for $x
eq16$). We get $\frac{\sqrt{x}-4}{x - 16}=\frac{\sqrt{x}-4}{(\sqrt{x}+4)(\sqrt{x}-4)}=\frac{1}{\sqrt{x}+4}$.

Step4: Evaluate the limit

Now, find $\lim_{x
ightarrow16}\frac{1}{\sqrt{x}+4}$. Substitute $x = 16$ into $\frac{1}{\sqrt{x}+4}$. When $x = 16$, $\sqrt{x}=4$, and $\frac{1}{\sqrt{x}+4}=\frac{1}{4 + 4}=\frac{1}{8}$.

Answer:

$\frac{1}{8}$