Limit of x^x as x approaches 0+

Answer: The limit of x^x as x approaches 0+ is equal to 1. That is,

limx→0+ xx = 1.

Limit of x^x when x tends to 0+

Question: Find the limit of x^x when x tends to 0+?

Solution

We need to find lim_x→0+ x^x.

Step 1:

Note that x = e^{ln x}

⇒ $x^x = e^{\ln(x^x)}$

⇒ $x^x = e^{x\ln x}$ as we know that ln(a^k) = kln(a).

Therefore, we have:

lim_x→0+ x^x

= lim_x→0+ $e^{x\ln x}$

⇒ lim_x→0+ x^x = $e^{\lim \limits_{x \to 0+} x\ln x}$ …(∗)

Step 2:

Now compute $\lim \limits_{x \to 0+} x\ln x$. It can be written as

$\lim \limits_{x \to 0+} \dfrac{\ln x}{1/x}$ (-∞/∞ form, so apply L’Hospital rule)

= $\lim \limits_{x \to 0+} \dfrac{(\ln x)’}{(1/x)’}$

= $\lim \limits_{x \to 0+} \dfrac{\frac{1}{x}}{-\frac{1}{x^2}}$

= $\lim \limits_{x \to 0+} (-x)$

= 0.

Step 3:

From (∗), we obtain that

lim_x→0+ x^x = $e^{0}$ = 1.

So the limit is 1.

Therefore, the limit of x^x is equal to 1 as x approaches 0+

Also Read: Limit of x sin(1/x) as x approaches 0

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