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

### Exercise

Evaluate the following limit:

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

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

### 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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