Question

In Problems 7– 42, find each limit algebraically.

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

Short answer

The answer is$\frac{2}{3}.$

Step-by-step solution

Step. 1 Given Information

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

​$f\left(x\right)=\frac{x^3-x^2+x-1}{x^4-x^3+2x-2}$

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

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

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

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

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

Step. 2 Factorizing

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

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

So,

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

Now we can put the limit directly.

Step. 3 Final calculation of the limit

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