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Question

Question: How do you find the limit of \[x\left( {{a^{\dfrac{1}{x}}} - 1} \right)\] as \[x \to \infty \]?...

How do you find the limit of x(a1x1)x\left( {{a^{\dfrac{1}{x}}} - 1} \right) as xx \to \infty ?

Explanation

Solution

Apply L'Hospital’s rule to evaluate the equation. Let us know the statement of L'Hospital’s rule: It states that the limit when we divide one function by another is the same after we take the derivative of each function.

Formula used:
limxcf(x)g(x)=limxcf(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)}}
As where the limit of x is till infinity.

Complete step by step solution:
Let us write the given input
x(a1x1)x\left( {{a^{\dfrac{1}{x}}} - 1} \right)
Rewriting with respect to limit as
limx(a1x11x)\mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{{a^{\dfrac{1}{x}}} - 1}}{{\dfrac{1}{x}}}} \right)
Hence, after simplifying it produces the indeterminate form 00  \begin{gathered} \dfrac{0}{0} \\\ \\\ \end{gathered}
Now apply L'Hospitals rule
limxcf(x)g(x)=limxcf(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)}}
limx(a1x11x)=limxa1xln(a)(1x2)1x2\mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{{a^{\dfrac{1}{x}}} - 1}}{{\dfrac{1}{x}}}} \right) = \mathop {\lim }\limits_{x \to \infty } \dfrac{{{a^{\dfrac{1}{x}}} \cdot \ln \left( a \right) \cdot \left( {\dfrac{{ - 1}}{{{x^2}}}} \right)}}{{\dfrac{{ - 1}}{{{x^2}}}}}
limx(a1x11x)=limxa1xln(a)\mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{{a^{\dfrac{1}{x}}} - 1}}{{\dfrac{1}{x}}}} \right) = \mathop {\lim }\limits_{x \to \infty } {a^{\dfrac{1}{x}}}\ln \left( a \right)
limx(a1x11x)=a0ln(a)\mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{{a^{\dfrac{1}{x}}} - 1}}{{\dfrac{1}{x}}}} \right) = {a^0}\ln \left( a \right)
Implies that
limx(a1x11x)=ln(a)\mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{{a^{\dfrac{1}{x}}} - 1}}{{\dfrac{1}{x}}}} \right) = \ln \left( a \right)
In which a0{a^0}is 1, hence the limit of
x(a1x1)=ln(a)x\left( {{a^{\dfrac{1}{x}}} - 1} \right) = \ln \left( a \right)
Or in terms of log, we get logea{\log _e}a.

Additional information:
Here are some of the properties to find the limit functions:
Sum Rule: This rule states that the limit of the sum of two functions is equal to the sum of their limits.
Constant Function Rule: The limit of a constant function is the constant.
Constant Multiple Rule: The limit of a constant times a function is equal to the product of the constant and the limit of the function.
Product Rule: This rule says that the limit of the product of two functions is the product of their limits (if they exist).
Quotient Rule: The limit of quotient of two functions is the quotient of their limits, provided that the limit in the denominator function is not zero

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.