How do you find the limit $lim_(t->-3)(t^2-9)/(2t^2+7t+3)$ ?

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Answer 1 · 2014-09-21T15:26:50

$lim_{t to -3}{t^2-9}/{2t^2+7t+3}$

by factoring out the numerator and the denominator,

$=lim_{t to -3}{(t+3)(t-3)}/{(t+3)(2t+1)}$

by cancelling out $(t-3)$ 's,

$=lim_{t to -3}{t-3}/{2t+1}={(-3)-3}/{2(-3)+1}={-6}/{-5}=6/5$

How do you find the limit lim_(t->-3)(t^2-9)/(2t^2+7t+3)lim_(t->-3)(t^2-9)/(2t^2+7t+3) ?