Question

In Problems 7– 42, find each limit algebraically.

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

Short answer

The answer is 4.

Step-by-step solution

Step. 1 Given Information

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

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

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

$x^4-1=1^4-1=1-1=0.$

Put in the denominator we get,

$x-1=1-1=0$.

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

Step. 2 Factorizing 

Denominator has only one factor i.e., $x-1$.

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

So,

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

Now we can put the limit directly.

Step. 3 Final calculation of the limit 

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