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.

22.\(\mathop {\lim }\limits_{x \to {a^ + }} \frac{{\cos x\ln (x - a)}}{{\ln \left( {{e^x} - {e^a}} \right)}}\)

Short answer

The value of \(\mathop {\lim }\limits_{x \to {a^ + }} \frac{{\cos x\ln (x - a)}}{{\ln \left( {{e^x} - {e^a}} \right)}}\) is \(\cos a\).

Step-by-step solution

Given information.

The given limit is \(\mathop {\lim }\limits_{x \to {a^ + }} \frac{{\cos x\ln (x - a)}}{{\ln \left( {{e^x} - {e^a}} \right)}}\).

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\( - \infty \)).

Find the value of the limit

Consider given limit function as product of two functions. \(\mathop {\lim }\limits_{x \to {a^ + }} \frac{{\cos x\ln (x - a)}}{{\ln \left( {{e^x} - {e^a}} \right)}} = \mathop {\lim }\limits_{x \to {a^ + }} \cos x \cdot \mathop {\lim }\limits_{x \to {a^ + }} \frac{{\ln (x - a)}}{{\ln \left( {{e^x} - {e^a}} \right)}}\) ...... (1)

Consider the limit function, \(\mathop {\lim }\limits_{x \to {a^ + }} \frac{{\ln (x - a)}}{{\ln \left( {{e^x} - {e^a}} \right)}}\).

Obtain the value of the function as \(x\) approaches \({a^ + }\).

As \(x\) approaches \({a^ + }\), the numerator is:

\(\begin{array}{c}\ln (x - a) = \ln (a - a)\\ = \ln (0)\\ = \infty \end{array}\)

And the value of denominator is:

\(\begin{array}{c}\ln \left( {{e^x} - {e^a}} \right) = \ln \left( {{e^a} - {e^a}} \right)\\ = \ln (0)\\ = \infty \end{array}\)

Thus, \(\mathop {\lim }\limits_{x \to {a^ + }} \frac{{\ln (x - a)}}{{\ln \left( {{e^x} - {e^a}} \right)}} = \frac{\infty }{\infty }\) is in an indeterminate form.

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

\(\begin{array}{c}\mathop {\lim }\limits_{x \to 1} \frac{{\ln (x - a)}}{{\ln \left( {{e^x} - {e^a}} \right)}} = \mathop {\lim }\limits_{x \to {a^ + }} \frac{{\frac{1}{{x - a}}}}{{\frac{{{e^x}}}{{{e^{x - {e^a}}}}}}}\\ = \mathop {\lim }\limits_{x \to {a^ + }} \frac{1}{{x - a}}\frac{{{e^x} - {e^a}}}{{{e^x}}}\\ = \mathop {\lim }\limits_{x \to {a^ + }} \frac{1}{{{e^x}}}\frac{{{e^x} - {e^a}}}{{x - a}}\\ = \mathop {\lim }\limits_{x \to {a^ + }} \frac{1}{{{e^x}}} \cdot \mathop {\lim }\limits_{x \to {a^ + }} \frac{{{e^x} - {e^a}}}{{x - a}}\end{array}\)

Therefore, \(\mathop {\lim }\limits_{x \to {a^ + }} \frac{{\ln (x - a)}}{{\ln \left( {{e^x} - {e^a}} \right)}} = \lim \frac{1}{{{e^x}}} \cdot \mathop {\lim }\limits_{x \to {a^ + }} \frac{{{e^x} - {e^a}}}{{x - a}}\). ...... (2)

Consider, \(\mathop {\lim }\limits_{x \to {a^ + }} \frac{{{e^x} - {e^a}}}{{x - a}}\).

Obtain the value of the function as \(x\) approaches \({a^ + }\).

As \(x\) approaches \({a^ + }\), the numerator is:

\(\begin{array}{c}{e^x} - {e^a} = {e^a} - {e^a}\\ = 0\end{array}\)

And the value of denominator is:

\(\begin{array}{c}x - a = a - a\\ = 0\end{array}\)

Thus, \(\mathop {\lim }\limits_{x \to {a^ + }} \frac{{{e^x} - {e^a}}}{{x - a}} = \frac{0}{0}\) is again in an indeterminate form.

Apply L'Hospital's Rule and obtain the limit

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

\(\begin{array}{c}\mathop {\lim }\limits_{x \to {a^ + }} \frac{{{e^x} - {e^a}}}{{x - a}} = \mathop {\lim }\limits_{x \to {a^ + }} \frac{{{e^x}}}{1}\\ = \mathop {\lim }\limits_{x \to {a^ + }} {e^x}\\ = {e^a}\end{array}\)

Substitute \(\mathop {\lim }\limits_{x \to {a^ + }} \frac{{{e^x} - {e^a}}}{{x - a}} = {e^a}\) in equation (2).

\(\begin{array}{c}\mathop {\lim }\limits_{x \to {a^ + }} \frac{{\ln (x - a)}}{{\ln \left( {{e^x} - {e^a}} \right)}} = \mathop {\lim }\limits_{x \to {a^ + }} \frac{1}{{{e^x}}} \cdot \mathop {\lim }\limits_{x \to {a^ + }} \frac{{{e^x} - {e^a}}}{{x - a}}\\ = \mathop {\lim }\limits_{x \to {a^ + }} \frac{1}{{{e^x}}} \cdot {e^a}\\ = 1\end{array}\)

Substitute the respective value in the equation (1).

\(\begin{array}{c}\mathop {\lim }\limits_{x \to {a^ + }} \frac{{\cos x\ln (x - a)}}{{\ln \left( {{e^x} - {e^a}} \right)}} = \mathop {\lim }\limits_{x \to {a^ + }} \cos x \cdot \mathop {\lim }\limits_{x \to {a^ + }} \frac{{\ln (x - a)}}{{\ln \left( {{e^x} - {e^a}} \right)}}\\ = \mathop {\lim }\limits_{x \to {a^ + }} \cos x \cdot 1\\ = \mathop {\lim }\limits_{x \to {a^ + }} \cos x\\ = \cos a\end{array}\)

Thus, the limit of the function is \(\cos a\).