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Q. 30

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Found in: Page 883

### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

# In Problems 7– 42, find each limit algebraically.$\underset{x\to -1}{\mathrm{lim}}\frac{{x}^{2}+x}{{x}^{2}-1}.$

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

See the 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

$\underset{x\to -1}{\mathrm{lim}}\frac{x}{x-1}=\frac{-1}{-1-1}\phantom{\rule{0ex}{0ex}}=\frac{-1}{-2}=\frac{1}{2}.$