Question

Evaluate the limit:
$\underset{x\to -1/2}{lim}\frac{8x^3+1}{2x+1}$

Answer 1

At $x\to -\frac{1}{2}$,

$\frac{8x^3+1}{2x+1}$ becomes $\frac{0}{0}$ form

Now,
$\underset{x\to -1/2}{lim}\frac{(2x{)}^3-(-1{)}^3}{2x-(-1)}$

$=\underset{x\to -1/2}{lim}\left[\frac{(2x{)}^3-(-1{)}^3}{2x-(-1)}\right]$

Taking $2x=y$

Also,
$x\to -1/2\Rightarrow 2x\to -1\Rightarrow y\to -1$

$=\underset{y\to -1}{lim}\frac{y^3-(-1{)}^3}{y-(-1)}$

$=3(-1{)}^{3-1}[\because \underset{x\to a}{lim}\frac{x^n-a^n}{x-a}=n{a}^{n-1}]$
$=3$