Question

1–38 ■ 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.

12.\(\mathop {lim}\limits_{t \to 0} \frac{{{8^t} - {5^t}}}{t}\).

Short answer

The value of \(\mathop {\lim }\limits_{t \to 0} \frac{{{8^t} - {5^t}}}{t}\)is\(\ln \left( {\frac{8}{5}} \right)\).

Step-by-step solution

Given limit

The given limit is \(\mathop {\lim }\limits_{t \to 0} \frac{{{8^t} - {5^t}}}{t}\).

Concept of L'Hospital's Rule

Suppose\(f\)and\(g\)are differentiable and\({g^\prime }(x) \ne 0\)near\(a\)(except possibly at\(a\)).

Suppose that\(\mathop {\lim }\limits_{x \to a} f(x) = 0\)and\(\mathop {\lim }\limits_{x \to a} g(x) = 0\)or that\(\mathop {\lim }\limits_{x \to a} f(x) = \pm \infty \)and\(\mathop {\lim }\limits_{x \to a} g(x) = \pm \infty \)or in other words, we have an indeterminate form of type\(\frac{0}{0}\)or\(\frac{\infty }{\infty }\).

Then,\(\mathop {\lim }\limits_{x \to a} \frac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to a} \frac{{{f^\prime }(x)}}{{{g^\prime }(x)}}\)if the limit on the right side exists (or is\(\infty \)or\( - \infty \)).

Evaluate the given limit and apply L’Hospital’s rule

Obtain the value of the function as \(t\) approaches \(0\).

As \(t\) approaches 0, the numerator is:

\(\begin{array}{c}{8^t} - {5^t} = {8^0} - {5^0}\\ = 1 - 1\\ = 0\end{array}\)

As \(t\) approaches 0, the denominator \(t\) is also \(0\).

Thus, \(\mathop {\lim }\limits_{t \to 0} \frac{{{8^t} - {5^t}}}{t} = \frac{0}{0}\) is in an indeterminate form.

Therefore, apply L'Hospital's Rule and obtain the limit.

\(\begin{array}{c}\mathop {\lim }\limits_{t \to 0} \frac{{{8^t} - {5^t}}}{t} = \mathop {\lim }\limits_{t \to 0} \frac{{{8^t}\ln 8 - {5^t}\ln 5}}{1}\\ = \frac{{{8^0}\ln 8 - {5^^\circ }\ln 5}}{1}\\ = \frac{{\ln 8 - \ln 5}}{1}\\ = \ln \left( {\frac{8}{5}} \right)\end{array}\)

Thus, the limit function is \(\ln \left( {\frac{8}{5}} \right)\).