# Calculating Limit of Function – Difference of functions to one – Exercise 6301

### Exercise

Evaluate the following limit:

$$\lim _ { x \rightarrow 1} (\frac{x}{x-1}-\frac{1}{\ln x})$$

$$\lim _ { x \rightarrow 1} (\frac{x}{x-1}-\frac{1}{\ln x})=\frac{1}{2}$$

### Solution

First, we try to plug in $$x = 1$$ and get

$$(\frac{1}{1-1}-\frac{1}{\ln 1})=$$

$$=(\frac{1}{0}-\frac{1}{0})=$$

$$=(\infty-\infty)$$

Note: The zero in the denominator is not absolute zero, but a number that is tending to zero.

We got the phrase $$\infty-\infty$$ (=infinity minus infinity). This is an indeterminate form, therefore we have to get out of this situation.

$$\lim _ { x \rightarrow 1} (\frac{x}{x-1}-\frac{1}{\ln x})=$$

In order to use Lopital Rule, we will break down the polynomials into factors and calculate the least common denominator

$$=\lim _ { x \rightarrow 1} \frac{x\ln x-(x-1)}{(x-1)\ln x}=$$

$$=\lim _ { x \rightarrow 1} \frac{x\ln x-x+1}{x\ln x-\ln x}=$$

We plug in one again and get

$$= \frac{1\cdot\ln 1-1+1}{1\cdot \ln 1-\ln 1}=$$

$$= \frac{0}{0}$$

We got the phrase $$\frac{"0"}{"0"}$$(=tending to zero divides tending to zero). This is also an indeterminate form, in such cases we use Lopital Rule – we derive the numerator and denominator separately and we will get

$$=\lim _ { x \rightarrow 1} \frac{\ln x+1-1}{\ln x+1-\frac{1}{x}}=$$

We simplify the expression and get

$$=\lim _ { x \rightarrow 1} \frac{\ln x}{\ln x+1-\frac{1}{x}}=$$

We plug in one again and get

$$=\frac{\ln 1}{\ln 1+1-\frac{1}{1}}=$$

$$=\frac{0}{0}$$

We got the phrase $$\frac{0}{0}$$ again, therefore we use Lopital Rule again – we derive the numerator and denominator separately and we will get

$$=\lim _ { x \rightarrow 1} \frac{\frac{1}{x}}{\frac{1}{x}+\frac{1}{x^2}}=$$

We simplify the expression and get

$$=\lim _ { x \rightarrow 1} \frac{x}{x+1}=$$

We plug in one again and this time we get

$$=\frac{1}{1+1}=$$

$$=\frac{1}{2}$$

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