Problem 2-4: Computing Limits of a Rational Function


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)}<br />
{x+3}=\displaystyle\lim_{x\to -3}{x-1}=-4

Problem 2-4

Published by Yaz

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