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

Exercise

Evaluate the following limit:

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

Final Answer

Show final answer

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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