How do you find $lim (5+6t^-1+t^-2)/(8-3t^-1+2t^-2)$ as $t->0$ ?

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Answer 1 · 2017-12-05T17:05:52

$1/2$

As the power is negative that it is hard to take the limit, we multiply by $t^2/t^2$ to cancel out the negative power.

$lim_(t->0) (5+6t^(-1)+t^(-2))/(8-t^(-1)+2t^(-2))* t^2/t^2$

$=lim_(t->0) (5t^2+6t+1)/(8t^2-t+2$

$=(5(0)^2+6(0)+1)/(8(0)^2-(0)+2$

$=1/2$

Here is the answer :)

How do you find lim (5+6t^-1+t^-2)/(8-3t^-1+2t^-2)lim (5+6t^-1+t^-2)/(8-3t^-1+2t^-2) as t->0t->0?