Question

Find the limit. Use L’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If L’Hospital’s Rule doesn’t apply, explain why.

15. \(\mathop {\lim }\limits_{t \to 0} \frac{{{e^{2t}} - 1}}{{\sin t}}\)

Short answer

The limit is \(\mathop {\lim }\limits_{t \to 0} \frac{{{e^{2t}} - 1}}{{\sin t}} = 2\).

Step-by-step solution

Apply L’Hospital’s Rule

TheL’Hospital’s Rulestates that for any real number\(a\), in case:

\(\mathop {\lim }\limits_{t \to a} \frac{{f\left( t \right)}}{{g\left( t \right)}} = \frac{0}{0},{\rm{ or, }}\mathop {\lim }\limits_{t \to a} \frac{{f\left( t \right)}}{{g\left( t \right)}} = \frac{\infty }{\infty }\),

Then the limit of the function can be calculated as:

\(\mathop {\lim }\limits_{t \to a} \frac{{f\left( t \right)}}{{g\left( t \right)}} = \mathop {\lim }\limits_{t \to a} \frac{{f'\left( t \right)}}{{g'\left( t \right)}}\)

Solving for given limit.

The given limit is:

\(\mathop {\lim }\limits_{t \to 0} \frac{{{e^{2t}} - 1}}{{\sin t}}\)

Here, we have \(f\left( x \right){\rm{ and }}g\left( x \right)\) both the functions are differentiable.

\(\mathop {\lim }\limits_{t \to 0} \frac{{{e^{2t}} - 1}}{{\sin t}} = \frac{0}{0}\)

So, by usingL’Hospital’s Rule, we have:

\(\begin{array}{c}\mathop {\lim }\limits_{t \to 0} \frac{{f\left( t \right)}}{{g\left( t \right)}} = \mathop {\lim }\limits_{t \to 0} \frac{{f'\left( t \right)}}{{g'\left( t \right)}}\\ = \mathop {\lim }\limits_{t \to 0} \frac{{2{e^{2t}}}}{{\cos t}}\\ = \frac{{2\left( 1 \right)}}{{\left( 1 \right)}} = 2\end{array}\)

Hence, the required value is \(\mathop {\lim }\limits_{t \to 0} \frac{{{e^{2t}} - 1}}{{\sin t}} = 2\).