Problem 2-4: Computing Limits of a Rational Function

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Compute
a) $\displaystyle\lim_{x\to 3}\frac{x^2+2x-3}{x+3}$
b) $\displaystyle\lim_{x\to -3}\frac{x^2+2x-3}{x+3}$

Solution

a) $\displaystyle\lim_{x\to 3}\frac{x^2+2x-3}{x+3}=\frac{3^2+2\times 3-3}{3+3}=2$

b) $\displaystyle\lim_{x\to -3}\frac{x^2+2x-3}{x+3}$
Since $\frac{(-3)^2+2\times (-3)-3}{(-3)+3}=\frac{0}{0}$ , we need to use an algebraic trick:

$\displaystyle\lim_{x\to -3}\frac{x^2+2x-3}{x+3}=\displaystyle\lim_{x\to -3}\frac{(x-1)(x+3)} {x+3}=\displaystyle\lim_{x\to -3}{x-1}=-4$

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