Question

In Problems 7– 42, find each limit algebraically.

$\underset{x\to 1}{\lim }\frac{x^3-x^2+3x-3}{x^2+3x-4}.$

Short answer

The answer is$\frac{4}{5}.$

Step-by-step solution

Step. 1 Given Information

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

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

Put $x=1$in the numerator we get,

$x^3-x^2+3x-3=1-1+3-3=0$

Put $x=1$in the denominator we get,

$x^2+3x-4=1+3-4=0.$

Since both numerator and denominator gives 0 means they both have $x=1$as their common root.

Step. 2 Factorizing 

Numerator , $x^3-x^2+3x-3=x^2(x-1)+3(x-1)=(x^2+3)(x-1).$

Denominator,$x^2+3x-4=x^2+4x-x-4=x(x+4)-1(x+4)=(x+4)(x-1).$

So,

$$\begin{gathered} \underset{x\to 1}{\lim }\frac{x^3-x^2+3x-3}{x^2+3x-4}=\underset{x\to 1}{\lim }\frac{(x^2+3)(x-1)}{(x+4)(x-1)} \\ =\underset{x\to 1}{\lim }\frac{x^2+3}{x+4}. \end{gathered}$$

Now we can put the limit directly.

Step. 3 Final calculation of the limit

$\underset{x\to 1}{\lim }\frac{x^2+3}{x+4}=\frac{1+3}{1+4}=\frac{4}{5}.$