Calculating Limit of Function- A function with ln in the power of x to 0 from right – Exercise 6323

Exercise

Evaluate the following limit:

limx0+(ln1x)x\lim _ { x \rightarrow 0^+} {(\ln \frac{1}{x})}^x

Final Answer


limx0+(ln1x)x=0\lim _ { x \rightarrow 0^+} {(\ln \frac{1}{x})}^x=0

Solution

First, we try to plug in x=0+x = 0^+ and get

(ln10+)0+{(\ln \frac{1}{0^+})}^{0^+}

We got the phrase 0\infty^0 (=infinity in the power of zero). This is an indeterminate form, therefore we have to get out of this situation.

limx0+(ln1x)x=\lim _ { x \rightarrow 0^+} {(\ln \frac{1}{x})}^x=

Using Logarithm Rules we get

=limx0+eln(ln1x)x==\lim _ { x \rightarrow 0^+} e^{\ln{(\ln \frac{1}{x})}^x}=

=limx0+exln(ln1x)==\lim _ { x \rightarrow 0^+} e^{x\ln(\ln \frac{1}{x})}=

We enter the limit inside:

=elimx0+xln(ln1x)== 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

=e0+ln(ln10+)== e^{ 0^+\cdot\ln(\ln \frac{1}{0^+})}=

=e0+ln(ln)== e^{ 0^+\cdot\ln(\ln \infty)}=

=e0+= e^{ 0^+\cdot\infty}

We got the phrase 00\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:

=elimx0+ln(ln1x)1x== e^{\lim _ { x \rightarrow 0^+} \frac{\ln(\ln \frac{1}{x})}{\frac{1}{x}}}=

We plug in 0+ again and get

=eln(ln10+)10+== e^{\frac{\ln(\ln \frac{1}{0^+})}{\frac{1}{0^+}}}=

=e== 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

=elimx0+1ln1x11x(1x2)1x2== 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

=elimx0+11xln1x== e^{\lim _ { x \rightarrow 0^+} \frac{1}{\frac{1}{x}\ln\frac{1}{x}}}=

We plug in 0+ again and this time we get

=110+ln10+== \frac{1}{\frac{1}{0^+}\ln\frac{1}{0^+}}=

=1ln== \frac{1}{\infty\cdot\ln\infty}=

=1== \frac{1}{\infty\cdot\infty}=

=1== \frac{1}{\infty}=

=0= 0

Note: A finite number divides by infinity is defined and equals to infinity. For the full list press here

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