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Question: How do you find the limit of \[{x^{\left( {\dfrac{1}{{\ln x}}} \right)}}\] as \[x\] approaches to 0 ...

How do you find the limit of x(1ln⁡x){x^{\left( {\dfrac{1}{{\ln x}}} \right)}} as xx approaches to 0 from positive side?

Explanation

Solution

Apply L'Hospital’s rule to evaluate the equation. Let us know the statement of L'Hospital’s rule defines: It states that the limit when we divide one function by another is the same after we take the derivative of each function and to evaluate the equation as we have an indeterminate form lim⁡x→0+\mathop {\lim }\limits_{x \to {0^ + }} we need to apply log to the function and simplify the terms.
It is denoted as:
lim⁡x→cf(x)g(x)=lim⁡x→cf′(x)g′(x)\mathop {\lim }\limits_{x \to c} \dfrac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to c} \dfrac{{f'(x)}}{{g'(x)}}

Complete step by step solution:
Let us write the given limit
x(1ln⁡x){x^{\left( {\dfrac{1}{{\ln x}}} \right)}}
Let us first plug in 0 right away, keeping in mind that we're approaching from the right side.
lim⁡x→0+1xln⁡x=10−∞=00\mathop {\lim }\limits_{x \to {0^ + }} \dfrac{1}{{{x^{\ln x}}}} = \dfrac{1}{{{0^{ - \infty }}}} = {0^0}
This is an indeterminate form, so we must use L'Hospital's rule.
This function does not appear to be in rational function form, f(x)g(x)\dfrac{{f\left( x \right)}}{{g\left( x \right)}}; however, a little manipulation will yield that form.
In fact, the following route is usually the one that should be taken when dealing with the indeterminate limit of a function to the power of a function:
Let,
y=x1ln⁡xy = {x^{\dfrac{1}{{\ln x}}}}
Then, apply logarithm to both sides we get
ln⁡y=ln⁡(x1ln⁡x)\ln y = \ln \left( {{x^{\dfrac{1}{{\ln x}}}}} \right)
ln⁡y=1ln⁡x(ln⁡x)\ln y = \dfrac{1}{{\ln x}}\left( {\ln x} \right)
ln⁡y=ln⁡(x)ln⁡(x)\ln y = \dfrac{{\ln \left( x \right)}}{{\ln \left( x \right)}}
ln⁡(xa)=aln⁡(x)\ln \left( {{x^a}} \right) = a\ln \left( x \right)
Simplifying the terms, we get
ln⁡y=1\ln y = 1
lim⁡x→0+ln⁡y=lim⁡x→0+1=1\mathop {\lim }\limits_{x \to {0^ + }} \ln y = \mathop {\lim }\limits_{x \to {0^ + }} 1 = 1
So, we have the limit for ln⁡y\ln y, but we want the limit for y as
lim⁡x→0+y=lim⁡x→0+eln⁡y=e1=e\mathop {\lim }\limits_{x \to {0^ + }} y = \mathop {\lim }\limits_{x \to {0^ + }} {e^{\ln y}} = {e^1} = e

Therefore, the limit of x(1ln⁡x){x^{\left( {\dfrac{1}{{\ln x}}} \right)}} as xx approaches to 0 from positive side is lim⁡x→0+1xln⁡x=e\mathop {\lim }\limits_{x \to {0^ + }} \dfrac{1}{{{x^{\ln x}}}} = e

Note:
For a limit approaching the given value, the original functions must be differentiable on either side of value, but not necessarily at the value given. The limit of a quotient is equal to the quotient of the limits. The limit of a constant function is equal to the constant. The limit of a linear function is equal to the number x is approaching.