Exercise
Evaluate the following limit:
\lim _ { x \rightarrow 1} x^{\frac{1}{1-x}}
Final Answer
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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