Question

use lhospital to determine the following limit. use exact values. \\(\lim\limits_{x\to 1} \left( \dfrac{\ln x}{5^{\ln x} - x} \
ight) = \\)

Explanation:

Step1: Check indeterminate form

When \(x = 1\), \(\ln(1)=0\), \(5^{\ln(1)}-1=5^{0}-1 = 0\). So \(\lim_{x
ightarrow1}\frac{\ln x}{5^{\ln x}-x}\) is in the \(\frac{0}{0}\) indeterminate form.

Step2: Apply L'Hospital's Rule

Differentiate the numerator and denominator.
The derivative of \(y = \ln x\) is \(y^\prime=\frac{1}{x}\).
Let \(u=\ln x\), then \(y = 5^{\ln x}-x=5^{u}-x\). Using the chain - rule, \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}-1\). Since \(\frac{dy}{du}=5^{u}\ln5\) and \(\frac{du}{dx}=\frac{1}{x}\), the derivative of \(5^{\ln x}-x\) is \(\frac{5^{\ln x}\ln5}{x}-1\).
So \(\lim_{x
ightarrow1}\frac{\ln x}{5^{\ln x}-x}=\lim_{x
ightarrow1}\frac{\frac{1}{x}}{\frac{5^{\ln x}\ln5}{x}-1}\).

Step3: Substitute \(x = 1\)

Substitute \(x = 1\) into \(\frac{\frac{1}{x}}{\frac{5^{\ln x}\ln5}{x}-1}\).
When \(x = 1\), we have \(\frac{\frac{1}{1}}{\frac{5^{0}\ln5}{1}-1}=\frac{1}{\ln5 - 1}\).

Answer:

\(\frac{1}{\ln5 - 1}\)