Question

use lhôpitals rule to find the following limit.
lim_{x
ightarrow3^{+}}left(\frac{1}{x - 3}-\frac{1}{ln(x - 2)}
ight)
lim_{x
ightarrow3^{+}}left(\frac{1}{x - 3}-\frac{1}{ln(x - 2)}
ight)=square\text{ (type an integer or a simplified fraction.)}

Explanation:

Step1: Combine the fractions

First, find a common - denominator to combine the two fractions:
\[

$$\begin{align*} \frac{1}{x - 3}-\frac{1}{\ln(x - 2)}&=\frac{\ln(x - 2)-(x - 3)}{(x - 3)\ln(x - 2)} \end{align*}$$

\]
As \(x
ightarrow3^{+}\), we have the indeterminate form \(\frac{0}{0}\).

Step2: Apply L'Hôpital's rule

Differentiate the numerator and the denominator.
The derivative of the numerator \(y_1=\ln(x - 2)-(x - 3)\) using the sum - rule and basic derivative formulas: \(y_1^\prime=\frac{1}{x - 2}-1=\frac{1-(x - 2)}{x - 2}=\frac{3 - x}{x - 2}\).
The derivative of the denominator \(y_2=(x - 3)\ln(x - 2)\) using the product - rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u=x - 3\), \(u^\prime = 1\), \(v=\ln(x - 2)\), \(v^\prime=\frac{1}{x - 2}\). So \(y_2^\prime=\ln(x - 2)+\frac{x - 3}{x - 2}\).
Now, \(\lim_{x
ightarrow3^{+}}\frac{\ln(x - 2)-(x - 3)}{(x - 3)\ln(x - 2)}=\lim_{x
ightarrow3^{+}}\frac{\frac{3 - x}{x - 2}}{\ln(x - 2)+\frac{x - 3}{x - 2}}\), which is still in the \(\frac{0}{0}\) form.

Step3: Apply L'Hôpital's rule again

Differentiate the new numerator and denominator.
The derivative of the new numerator \(y_3 = 3 - x\) is \(y_3^\prime=-1\), and the derivative of the new denominator \(y_4=\ln(x - 2)+\frac{x - 3}{x - 2}=\ln(x - 2)+1-\frac{1}{x - 2}\).
\(y_4^\prime=\frac{1}{x - 2}+\frac{1}{(x - 2)^2}\).
Then \(\lim_{x
ightarrow3^{+}}\frac{-1}{\frac{1}{x - 2}+\frac{1}{(x - 2)^2}}\).

Step4: Evaluate the limit

Substitute \(x = 3\) into \(\frac{-1}{\frac{1}{x - 2}+\frac{1}{(x - 2)^2}}\):
\[

$$\begin{align*} \lim_{x ightarrow3^{+}}\frac{-1}{\frac{1}{x - 2}+\frac{1}{(x - 2)^2}}&=\frac{-1}{1 + 1}\\ &=-\frac{1}{2} \end{align*}$$

\]

Answer:

\(-\frac{1}{2}\)