Exercise
Evaluate the following limit:
\lim _ { x \rightarrow 0^+} {(\ln \frac{1}{x})}^x
Final Answer
Solution
First, we try to plug in x = 0^+ and get
{(\ln \frac{1}{0^+})}^{0^+}
We got the phrase \infty^0 (=infinity in the power of zero). This is an indeterminate form, therefore we have to get out of this situation.
\lim _ { x \rightarrow 0^+} {(\ln \frac{1}{x})}^x=
Using Logarithm Rules we get
=\lim _ { x \rightarrow 0^+} e^{\ln{(\ln \frac{1}{x})}^x}=
=\lim _ { x \rightarrow 0^+} e^{x\ln(\ln \frac{1}{x})}=
We enter the limit inside:
= e^{\lim _ { x \rightarrow 0^+} x\ln(\ln \frac{1}{x})}=
Note: This can be done because an exponential function is a continuous function.
We plug in 0+ again and get
= e^{ 0^+\cdot\ln(\ln \frac{1}{0^+})}=
= e^{ 0^+\cdot\ln(\ln \infty)}=
= e^{ 0^+\cdot\infty}
We got the phrase 0\cdot\infty(=tending to zero multiples by infinity). This is also an indeterminate form, it such cases we use Lopital Rule – we derive the numerator and denominator separately and we will get
In order to use Lopital Rule, we simplify the phrase and get:
= e^{\lim _ { x \rightarrow 0^+} \frac{\ln(\ln \frac{1}{x})}{\frac{1}{x}}}=
We plug in 0+ again and get
= e^{\frac{\ln(\ln \frac{1}{0^+})}{\frac{1}{0^+}}}=
= e^{\frac{\infty}{\infty}}=
We got the phrase \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 will get
= e^{\lim _ { x \rightarrow 0^+} \frac{\frac{1}{\ln\frac{1}{x}}\cdot \frac{1}{\frac{1}{x}}\cdot (-\frac{1}{x^2})}{-\frac{1}{x^2}}}=
We simplify the phrase and get
= e^{\lim _ { x \rightarrow 0^+} \frac{1}{\frac{1}{x}\ln\frac{1}{x}}}=
We plug in 0+ again and this time we get
= \frac{1}{\frac{1}{0^+}\ln\frac{1}{0^+}}=
= \frac{1}{\infty\cdot\ln\infty}=
= \frac{1}{\infty\cdot\infty}=
= \frac{1}{\infty}=
= 0
Note: A finite number divides by infinity is defined and equals to infinity. For the full list press here
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