Exercise
Evaluate the following limit:
\lim _ { x \rightarrow 1^-}(\ln x)(\ln(1-x))
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
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