Question

In Problems 7– 42, find each limit algebraically.

$\underset{x\to -1}{\lim }\frac{{\left(x+1\right)}^2}{x^2-1}.$

Short answer

The answer is 0.

Step-by-step solution

Step. 1 Given Information

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

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

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

${\left(x+1\right)}^2={\left(-1+1\right)}^2=0$.

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

$x^2-1={\left(-1\right)}^2-1=1-1=0$.

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

Step. 2 Factorizing 

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

Numerator ,${\left(x+1\right)}^2=(x+1)(x+1).$

So,

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

Now we can put the limit directly.

Step. 3 Final calculation of the limit 

$\underset{x\to -1}{\lim }\frac{x+1}{x-1}=\frac{-1+1}{-1-1}=\frac{0}{-2}=0.$