# Find the limit: \lim_{x\rightarrow -1} \frac{2x^2-x-3}{x+1}

## Question:

Find the limit: {eq}\lim_{x\rightarrow -1} \frac{2x^2-x-3}{x+1} {/eq}

## Limits; Rational Expressions:

We recall that if a polynomial {eq}p(x) {/eq} satisfies {eq}p(a)=0 {/eq} for some real number {eq}a {/eq}, then {eq}x {/eq} is divisible by the binomial {eq}x-a. {/eq} Using such factorization we can simplify the expressions in the numerator and denominator to evaluate the limit quickly.

## Answer and Explanation:

We consider the rational expression {eq}\frac{2x^2-x-3}{x+1} {/eq} which is defined for {eq}x\neq-1 {/eq} and notice that

{eq}\lim_{x\rightarrow -1} \frac{2x^2-x-3}{x+1}\text{ has indefinite form }\frac{0}{0}, {/eq}

so we can divide both numerator and denominator by {eq}x+1. {/eq} in fact we have:

{eq}\begin{align*} \frac{2x^2-x-3}{x+1}&= \frac{2x^2+2x-3x-3}{x+1}\\ &=\frac{2x(x+1)-3(x+1)}{x+1} \\ &=\frac{(2x-3)(x+1)}{x+1} \\ &=2x-3 \text{ for any }x\neq -1. \end{align*} {/eq}

So we have that:

{eq}\begin{align*} \lim_{x\rightarrow -1} \frac{2x^2-x-3}{x+1}&=\lim_{x\rightarrow -1} 2x-3\\ &=2(-1)-3=-5. \end{align*} {/eq}

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