Calculating Limit of Function – A multiplication of functions with ln one-sided to 1- – Exercise 6290

Exercise

Evaluate the following limit:

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

Final Answer

Show final answer

Solution

First, we try to plug in $x = 1^-$

When we aim for one from the left, we are close to one, but are smaller than one (e.g., 0.9999), therefore we get

$(\ln 1^-)(\ln(1-1^-))=0\cdot (-\infty)$

We got the phrase $0\cdot (-\infty)$ (=tends to zero multiplies by minus infinity). This is an indeterminate form, therefore we have to get out of this situation.

Note: The zero in the expression is not absolute, but a number tending to zero. Therefore, the expression is not equal to zero.

In order to use Lopital Rule, we simplify the expression and get

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

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

Now, if we plug in one in the phrase. we will get $\frac{\infty}{\infty}$ (=infinity divides by infinity). This is also an indeterminate form, in such cases, we use Lopital Rule – we derive the numerator and denominator separately and we get

$=\lim _ { x \rightarrow 1^-}\frac{\frac{-1}{1-x}}{-\frac{1}{\ln^2 x}\cdot \frac{1}{x}}=$

We simplify the expression

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

We plug in one again and get

$=\frac{1^-\ln^2 1^-}{1-1^-}=$

$\frac{0}{0}$

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

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

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

We simplify the expression

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

We plug in one again and this time we get

$=\frac{\ln^2 1^-+ 2\ln 1^-}{-1}=$

$=\frac{0+0}{-1}=$

$=0$

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!