Harold Kessler

2021-12-27

Let $f\left(x\right)=\frac{{\left(x+8\right)}^{2}}{{x}^{2}-64}$. Find a) $\underset{x⇒-8}{lim}f\left(x\right)$, b) $\underset{x⇒0}{lim}f\left(x\right)$, and c) $\underset{x⇒8}{lim}f\left(x\right)$.

hysgubwyri3

Step 1
a) Plug the given expression for $f\left(x\right)$, then factor the denominator and simplify.
$\underset{x⇒-8}{lim}f\left(x\right)$
$=\underset{x⇒-8}{lim}\frac{{\left(x+8\right)}^{2}}{{x}^{2}-64}$
$=\underset{x⇒-8}{lim}\frac{{\left(x+8\right)}^{2}}{{x}^{2}-{8}^{2}}$
$=\underset{x⇒-8}{lim}\frac{{\left(x+8\right)}^{2}}{\left(x+8\right)\left(x-8\right)}$
$=\underset{x⇒-8}{lim}\frac{\left(x+8\right)}{\left(x-8\right)}$
Step 2
Plug $x=-8$ for the limit
$=\underset{x⇒-8}{lim}\frac{\left(x+8\right)}{\left(x-8\right)}$
$=\frac{-8+8}{-8-8}$
$=\frac{0}{-16}$
$=0$

eninsala06

Step 3
b) Plug expression of $f\left(x\right)$ in the limit then simplify and plug $x=0$
$\underset{x⇒0}{lim}f\left(x\right)$
$=\underset{x⇒0}{lim}\frac{{\left(x+8\right)}^{2}}{{x}^{2}-64}$
$=\underset{x⇒0}{lim}\frac{{\left(x+8\right)}^{2}}{{x}^{2}-{8}^{2}}$
$=\underset{x⇒0}{lim}\frac{{\left(x+8\right)}^{2}}{\left(x+8\right)\left(x-8\right)}$
$=\underset{x⇒0}{lim}\frac{\left(x+8\right)}{\left(x-8\right)}$
$=\frac{0+8}{0-8}$
$=-1$

karton

Step 4
c) We simplify the limit same way and plug x=8 this time.

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