# Question #6cdd8

Oct 24, 2017

the answer is $a = - 250$

#### Explanation:

the given question is ${\lim}_{t \rightarrow 0} \frac{a t - \tan \left(5 t\right)}{{t}^{3}} = - \frac{125}{3}$
on applying L'Hospital's Rule we get
${\lim}_{t \rightarrow 0} \frac{a - 5 {\sec}^{2} \left(5 t\right)}{3 {t}^{2}} = - \frac{125}{3}$
since it is in indeterminate form again applying L'Hospital's Rule
${\lim}_{t \rightarrow 0} \frac{a - 50 \tan \left(5 t\right) {\sec}^{2} \left(5 t\right)}{6 t} = - \frac{125}{3}$
on apllying L'Hospital's Rule again
${\lim}_{t \rightarrow 0} \frac{a - 500 {\sec}^{4} \left(5 t\right) \tan \left(5 t\right)}{6} = - \frac{125}{3}$
on applying limit and solving we get $a = - 250$