fbpx
calculus online - Free exercises and solutions to help you succeed!

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


\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}

Have a question? Found a mistake? – Write a comment below!
Was it helpful? You can buy me a cup of coffee here, which will make me very happy and will help me upload more solutions! 

Share with Friends

Leave a Reply