Question

use lhospital to determine the following limit. use exact values. \\(\lim_{x\to 1} \left( \frac{\ln x}{\ln x - x} \
ight) = \\)

Explanation:

Step1: Check the form of the limit

When \(x = 1\), \(\ln(1)=0\), \(g^{\ln(1)}-1=g^{0}-1 = 1 - 1=0\). So, \(\lim_{x
ightarrow1}\frac{\ln x}{g^{\ln x}-x}\) is in the \(\frac{0}{0}\) 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 \(y = g^{u}-x\). Using the chain - rule, \(\frac{dy}{dx}=\frac{d(g^{u})}{du}\cdot\frac{du}{dx}-1\). Since \(\frac{d(g^{u})}{du}=g^{u}\ln g\) and \(\frac{du}{dx}=\frac{1}{x}\), the derivative of \(g^{\ln x}-x\) is \(\frac{g^{\ln x}\ln g}{x}-1\).
So, \(\lim_{x
ightarrow1}\frac{\ln x}{g^{\ln x}-x}=\lim_{x
ightarrow1}\frac{\frac{1}{x}}{\frac{g^{\ln x}\ln g}{x}-1}\).

Step3: Evaluate the new limit

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

Answer:

\(\frac{1}{\ln g - 1}\)