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How do you find $lim (1-5t^-1)/(4+6t^-1)$ as $t->0$ ?

1 Answer

Feb 11, 2017

$5/6$

Explanation:

Since $t^-1=1/t$ you could rewrite:

$lim_(t->0) (1-5/t)/(4-6/t)=lim_(t->0) ((t-5)/cancelt)/((4t-6)/cancelt)=lim_(t->0)(t-5)/(4t-6)=5/6$

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