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

### Exercise

Evaluate the following limit:

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

$$\lim _ { x \rightarrow 1^-}(\ln x)(\ln(1-x))=0$$

### 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

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

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

$$=0$$

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