Question

In Problems 7– 42, find each limit algebraically.

$\underset{x\to 2}{\lim }\frac{x^3-8}{x^2-4}.$

Short answer

The answer is 3.

Step-by-step solution

Step. 1 Given Information

Firstly, we check whether the given function is in indeterminant form or not.

$f\left(x\right)=\frac{x^3-8}{x^2-4}$

Put $x=2$in the numerator we get,

$x^3-8=2^3-8=8-8=0$.

Put $x=2$in the denominator we get,

$x^2-4=2^2-4=4-4=0$.

Since both numerator and denominator gives 0 means they both have $x=2$as their common root.

Step. 2 Factorizing 

Denominator,$x^2-4=(x+2)(x-2).$

Numerator ,$x^3-8=(x-2)(x^2+2x+4).$

So,

$$\begin{gathered} \underset{x\to 2}{\lim }\frac{x^3-8}{x^2-4}=\underset{x\to 2}{\lim }\frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)} \\ =\underset{x\to 2}{\lim }\frac{x^2+2x+4}{x+2}. \end{gathered}$$

Now we can put the limit directly.

Step. 3 Final calculation of the limit  

$$\begin{gathered} \underset{x\to 2}{\lim }\frac{x^2+2x+4}{x+2}=\frac{4+4+4}{2+2} \\ =\frac{12}{4}=3. \end{gathered}$$