Question

In Problems 7– 42, find each limit algebraically.

$\underset{x\to 2}{\lim }\frac{x^3-2x^2+4x-8}{x^2+x-6}.$

Short answer

The answer is $\frac{8}{5}.$

Step-by-step solution

Step. 1 Given Information

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

$f\left(x\right)=\frac{x^3-2x^2+4x-8}{x^2+x-6}$

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

$x^3-2x^2+4x-8=2^3-2\left(4\right)+4\left(2\right)-8=8-8+8-8=0.$

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

$x^2+x-6=4+2-6=0$

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

Step. 2 Factorizing 

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

Denominator,$x^2+x-6=x^2+3x-2x-6=x(x+3)-2(x+3)=(x+3)(x-2).$

So,

$$\begin{gathered} \underset{x\to 2}{\lim }\frac{x^3-2x^2+4x-8}{x^2+x-6}=\underset{x\to 2}{\lim }\frac{(x^2+4)(x-2)}{(x+3)(x-2)} \\ =\underset{x\to 2}{\lim }\frac{(x^2+4)}{(x+3)}. \end{gathered}$$

Now we can put the limit directly.

Step. 3 Final calculation of the limit

$\underset{x\to 2}{\lim }\frac{x^2+4}{x+3}=\frac{4+4}{2+3}=\frac{8}{5}.$