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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


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

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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