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QuestionFind the limit: lim⁡x→0+xx\lim _{x \rightarrow 0^{+}} x^{x}.

Studdy Solution

STEP 1

Assumptions1. We are given the limit as xx approaches 00 from the right, denoted as 0+0^{+}. . The function is xxx^{x}.
3. We are asked to find the value of this limit.

STEP 2

This limit is of the form 000^{0}, which is an indeterminate form. To solve this, we can use the logarithmic properties and the L'Hopital's Rule. First, we will take the natural logarithm (ln) of the function.
ln⁡(y)=ln⁡(xx)\ln(y) = \ln(x^{x})

STEP 3

Using the properties of logarithms, we can bring the exponent down.
ln⁡(y)=xln⁡(x)\ln(y) = x \ln(x)

STEP 4

Now, we will find the limit of the new function as xx approaches 0+0^{+}.
lim⁡x→0+ln⁡(y)=lim⁡x→0+xln⁡(x)\lim{x \rightarrow0^{+}} \ln(y) = \lim{x \rightarrow0^{+}} x \ln(x)

STEP 5

This limit is still of an indeterminate form, 0×−∞0 \times -\infty. To solve this, we can rewrite the function as a fraction and then apply L'Hopital's Rule.
lim⁡x→0+ln⁡(y)=lim⁡x→0+ln⁡(x)1/x\lim{x \rightarrow0^{+}} \ln(y) = \lim{x \rightarrow0^{+}} \frac{\ln(x)}{1/x}

STEP 6

Now, we can apply L'Hopital's Rule, which states that the limit of a quotient of two functions as xx approaches a certain value is equal to the limit of the quotients of their derivatives.
lim⁡x→0+ln⁡(y)=lim⁡x→0+ddxln⁡(x)ddx(1/x)\lim{x \rightarrow0^{+}} \ln(y) = \lim{x \rightarrow0^{+}} \frac{\frac{d}{dx} \ln(x)}{\frac{d}{dx} (1/x)}

STEP 7

Calculate the derivatives of the numerator and the denominator.
ddxln⁡(x)=1x\frac{d}{dx} \ln(x) = \frac{1}{x}ddx(1/x)=−1x2\frac{d}{dx} (1/x) = -\frac{1}{x^{2}}

STEP 8

Substitute the derivatives back into the limit.
lim⁡x→0+ln⁡(y)=lim⁡x→0+1x−1x2\lim{x \rightarrow0^{+}} \ln(y) = \lim{x \rightarrow0^{+}} \frac{\frac{1}{x}}{-\frac{1}{x^{2}}}

STEP 9

implify the fraction.
lim⁡x→+ln⁡(y)=lim⁡x→+−x\lim{x \rightarrow^{+}} \ln(y) = \lim{x \rightarrow^{+}} -x

STEP 10

Now, we can find the limit as xx approaches 0+0^{+}.
lim⁡x→0+ln⁡(y)=lim⁡x→0+−x=0\lim{x \rightarrow0^{+}} \ln(y) = \lim{x \rightarrow0^{+}} -x =0

STEP 11

Since we took the natural logarithm of the function at the beginning, we need to take the exponential of both sides to get the original function back.
y=elim⁡x→0+ln⁡(y)=e0y = e^{\lim{x \rightarrow0^{+}} \ln(y)} = e^{0}

STEP 12

Calculate the exponential.
y=e0=y = e^{0} =So, lim⁡x→0+xx=\lim{x \rightarrow0^{+}} x^{x} =.

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