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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


\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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