How do you find $lim (1+2t^-1)/(7+t^-1-5t^-2)$ as $t->0$ ?

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Answer 1 · 2017-11-02T18:47:42

The answer is $0$

$lim_(t->0)(1+2/t)/(7+1/t-5/t^2)=oo/oo$

This is an indeterminate form

Let's perform some simplifications

$lim_(t->0)(1+2/t)/(7+1/t-5/t^2)=lim_(x->0)((t+2)/t)/((7t^2+t-5)/t^2)$

$=lim_(x->0)((t+2)t)/((7t^2+t-5))$

$=0/-5=0$

graph{(1+2/x)/(7+1/x-5/x^2) [-7.02, 7.024, -3.51, 3.51]}

How do you find lim (1+2t^-1)/(7+t^-1-5t^-2)lim (1+2t^-1)/(7+t^-1-5t^-2) as t->0t->0?