Question

In Problems 7– 42, find each limit algebraically.

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

Short answer

The answer is$\frac{1}{2}.$

Step-by-step solution

Step. 1 Given Information

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

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

Since if we put $x=-1$we get,

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

So the limit is in 0/0 indeterminant form.

Step. 2 Try to find factor which make the function indeterminant.

Since we have both polynomial functions in the numerator and denominator, So let's factorize them,

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

cancelling $\left(x+1\right)$from both numerator and denominator we get,

$\frac{x}{x-1}$.

Now if we put $x=-1$then it is not in indeterminant from. So, now we can directly put the limit in it.

Step. 3 Solving the limit 

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