Chapter 4, Problem 33RE

To determine

Tofind:the limit of the function.

Answer to Problem 33RE

Thelimit of the function $\underset{\text{x}\to {\text{1}}^{\text{+}}}{\text{lim}}\left(\frac{\text{x}}{\text{x-1}}\text{-}\frac{\text{1}}{\text{ln}\left(\text{x}\right)}\right)$ is $\frac{1}{2}.$

Explanation of Solution

Given:

  $\underset{\text{x}\to {\text{1}}^{\text{+}}}{\text{lim}}\left(\frac{\text{x}}{\text{x-1}}\text{-}\frac{\text{1}}{\text{ln}\left(\text{x}\right)}\right)$ .

Concept used:

If the function will be in the form of indeterminate $\left(\frac{0}{0}\right)$ which is not valid.

In this kind of situation $\text{L}$ Hospital’s Rule.

Which state that the limit of a quotient of the functions is equal to the limit of quotient of the derivative.

Calculation:

  $\underset{\text{x}\to {\text{1}}^{\text{+}}}{\text{lim}}\left(\frac{\text{x}}{\text{x-1}}\text{-}\frac{\text{1}}{\text{ln}\left(\text{x}\right)}\right)$

The function can be written as:

  $\underset{\text{x}\to {\text{1}}^{\text{+}}}{\text{lim}}\left(1+\frac{1}{\text{x-1}}\text{-}\frac{\text{1}}{\text{ln}\left(\text{x}\right)}\right)$ .

  $\underset{\text{x}\to {\text{1}}^{\text{+}}}{\text{lim}}\left(\text{1+}\frac{\text{ln}\left(\text{x}\right)\text{-x+1}}{\left(\text{x-1}\right)\text{ln}\left(\text{x}\right)}\right)$ .

  $1+\underset{\text{x}\to {\text{1}}^{\text{+}}}{\text{lim}}\left(\frac{\text{ln}\left(\text{x}\right)\text{-x+1}}{\left(\text{x-1}\right)\text{ln}\left(\text{x}\right)}\right)$ .

By putting direct $x=0$ the function will be in the form of indeterminate $\left(\frac{0}{0}\right)$ which is not valid.

In this kind of situation $\text{L}$ Hospital’s Rule.

Which state that the limit of a quotient of the functions is equal to the limit of quotient of there derivative.

  $1+\underset{\text{x}\to {\text{1}}^{\text{+}}}{\text{lim}}\left(\frac{\text{ln}\left(\text{x}\right)\text{-x+1}}{\left(\text{x-1}\right)\text{ln}\left(\text{x}\right)}\right)$

  $1+\underset{\text{x}\to {\text{1}}^{\text{+}}}{\text{lim}}\left(\frac{\frac{d}{dx}\left(\text{ln}\left(\text{x}\right)\text{-x+1}\right)}{\frac{d}{dx}\left(\left(\text{x-1}\right)\text{ln}\left(\text{x}\right)\right)}\right)$ .

Finding derivative of numerator and denominator.

  $1+\underset{\text{x}\to {\text{1}}^{\text{+}}}{\text{lim}}\left(\frac{\frac{1}{x}-1}{1+\ln \left(x\right)-\frac{1}{x}}\right)$ .

By putting direct $x=0$ again the function will be in the form of indeterminate $\left(\frac{0}{0}\right)$ which is not valid.

this kind of situation $\text{L}$ Hospital’s Rule.

  $1+\underset{\text{x}\to {\text{1}}^{\text{+}}}{\text{lim}}\left(\frac{\frac{d}{dx}\left(\frac{1}{x}-1\right)}{\frac{d}{dx}\left(1+\ln \left(x\right)-\frac{1}{x}\right)}\right).$

  $1+\underset{\text{x}\to {\text{1}}^{\text{+}}}{\text{lim}}\left(\frac{-\frac{1}{x^2}}{\frac{1}{x}+\frac{1}{x^2}}\right).$

By putting it limits:

  $1+\left(\frac{-1}{\frac{1}{1}+\frac{1}{1^2}}\right).$

  $\underset{\text{x}\to {\text{1}}^{\text{+}}}{\text{lim}}\left(\frac{\text{x}}{\text{x-1}}\text{-}\frac{\text{1}}{\text{ln}\left(\text{x}\right)}\right)$$=$$\frac{1}{2}.$

Hence the limit of the function $\underset{\text{x}\to {\text{1}}^{\text{+}}}{\text{lim}}\left(\frac{\text{x}}{\text{x-1}}\text{-}\frac{\text{1}}{\text{ln}\left(\text{x}\right)}\right)$ is $\frac{1}{2}.$