Question

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

Answer 1

`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`