Question

Evaluate :-
Lim(t-> 3 ) [ 3+6t/(4t²-1)]

Answer 1

Hey there!

$\lim_{x \to 3} [\frac { 3+6t} {4t^2-1}]$
= $\lim_{x \to 3} [\frac { 3(1+2t)} { ( 2t+1)(2t-1)}]$
= $\lim_{x \to 3} (\frac{3} {2t-1})$
= $\frac {3} {2(3)-1}$
=$\frac {3} {5}$

Hope helped!

Answer 2

Hi,

Please see the attached file !

Thanks