Calculating Limit of Function – x in the power of a rational function to 1 – Exercise 6297

Exercise

Evaluate the following limit:

$\lim _ { x \rightarrow 1} x^{\frac{1}{1-x}}$

Final Answer

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Solution

First, we try to plug in $x = 1$ and get

$x^{\frac{1}{1-1}}$

We got the phrase ${"1"}^{"0"}$ (=tending to one in the power of tending to zero). This is an indeterminate form, therefore we have to get out of this situation.

$\lim _ { x \rightarrow 1} x^{\frac{1}{1-x}}=$

Using Logarithm Rules we get

$=\lim _ { x \rightarrow 1} e^{\ln x^{\frac{1}{1-x}}}=$

$=\lim _ { x \rightarrow 1} e^{\frac{1}{1-x}\cdot \ln x}=$

We simplify the expression and get

$=\lim _ { x \rightarrow 1} e^{\frac{\ln x}{1-x}}=$

We enter the limit inside and get

$=e^{\lim _ { x \rightarrow 1} \frac{\ln x}{1-x}}=$

Note: This can be done because an exponential function is a continuous function.

We plug in one again and get

$=e^{ \frac{\ln 1}{1-1}}=$

$=e^{ \frac{0}{0}}=$

We got the phrase $\frac{"0"}{"0"}$(=tending to zero divides tending to zero). This is also an indeterminate form, in such cases we use Lopital Rule – we derive the numerator and denominator separately and we will get

$=e^{\lim _ { x \rightarrow 1} \frac{\frac{1}{x}}{-1}}=$

We simplify the expression and get

$=e^{\lim _ { x \rightarrow 1} -\frac{1}{x}}=$

We plug in one again and this time we get

$=e^{ -\frac{1}{1}}=$

$=e^{-1}$

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