Solving limit shape $0^0$

Question

I have an exercise which is $\lim_{x \to 0^+} x^x$.

I have done it, and the answer is $1$, but I just need to know the easiest way to solve it without l'Hospital's rule, since in my country they don't allow us to use L'Hospital's rule.

Answer 1

Note that for $0 < x < 1$ we have the inequality $1-x < -\ln x < (1-x)/x.$

Hence,

$$2x (1- \sqrt{x}) < -x\ln x = -2x\ln \sqrt{x} < \frac{2x(1-\sqrt{x})}{\sqrt{x}}.$$

Applying the squeeze theorem we have $\lim_{x \to 0} x \ln x = 0$ and using the continuity of the exponential function

$$\lim_{x \to 0}x^x = \lim_{x \to 0}\exp(\ln x^x)=\lim_{x \to 0}\exp(x\ln x)=\exp(\lim_{x \to 0}x \ln x)= 1.$$

Answer 2

METHODOLOGY $1$

Here is one way to evaluate the limit without use of L'Hospital's Rule. We make use of the inequality

$$e^x\ge \left(1+\frac xn\right)^n \tag 1$$

for $x>-n$, which I proved in THIS ANSWER using only the limit definition of the exponential function along with Bernoulli's Inequality. Note that setting $n=2$ in $(1)$ reveals

$$e^x\ge 1+x+\frac14x^2 \tag 2$$

for $x>-2$.

Proceeding with the evaluation of the limit $\lim_{x\to 0^+}x^x$, we note that we can write $x^x=e^{x\log(x)}$. Next, we have

$$\begin{align} \lim_{x\to 0^+}x\log(x)&=-\lim_{x\to \infty}xe^{-x}\\\\ &=-\lim_{x\to \infty}\frac{x}{e^x} \end{align}$$

From $(2)$, we find that the term $\frac{x}{e^x}$ satisfies the inequalities

$$0\le \frac{x}{e^x}\le \frac{x}{1+x+\frac14x^2} \tag 3$$

whereupon applying the squeeze theorem to $(3)$ reveals

$$\lim_{x\to \infty}\frac{x}{e^x}=0$$

Finally, since $e^0=1$, we find the limit of interest is indeed

$$\lim_{x\to 0+}x^x=1$$

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METHODOLOGY $2$

Alternatively, I showed in THIS ANSWER and THIS ONE, using only the limit definition of the exponential function along with Bernoulli's Inequality, that the logarithm function satisfies the inequalities

$$\frac{z-1}{z}\le \log(z)\le z-1 \tag 4$$

Setting $z=x^\mu$, $\mu<1$ in $(4)$ yields

$$\frac{x^\mu -1}{\mu x^\mu}\le \log(x)\le \frac{x^\mu -1}{\mu}$$

Therefore, we have

$$\frac{x(1-x^{-\mu})}{\mu}\le x\log(x)\le \frac{x(x^\mu -1)}{\mu} \tag 5$$

whereupon applying the squeeze theorem to $(5)$ yields

$$\lim_{x\to 0^+}x\log(x)=0$$

and therefore

$$\lim_{x\to 0^+}x^x$$

Answer 3

Whew, this was harder than I thought. Hopefully someone will provide a shorter answer. Instead of showing that $x^x$ tends to $1$, you can instead show that $\ln(x^x) = x \ln(x)$ tends to $0$ as $x$ goes to $0$ from the right.

The derivative of $x \ln x$ is $h(x) = \ln x + 1$, and so for $x > 0$, $$ x \ln x = \int_1^x (\ln t + 1)dt - 1 \cdot \ln(1) = \int_1^x (\ln t + 1)dt$$ Now, you need to show that the area under the curve from $0$ to $1$ of $h(x)$ is $0$. If you draw a picture, you can instead consider the area vertically rather than horizontally, by integrating the inverse of $h$, which is $g(x) = e^{x-1}$.

Geometrically, you can see that $$\int_0^{\frac{1}{e}} (\ln t + 1)dt = -\int_{-\infty}^0 e^{t-1}dt = -\frac{1}{e}$$ and similarly $$1 = \int_{\frac{1}{e}}^1 (\ln t + 1)dt + \int_0^1 e^{t-1}dt = \int_{\frac{1}{e}}^1 (\ln t + 1)dt + \frac{e-1}{e}$$ so $\int_{1/e}^1 (\ln t + 1)dt = 1 - \frac{e-1}{e} = \frac{1}{e}$. Thus $$\int_0^1 (\ln t + 1)dt = -\frac{1}{e} + \frac{1}{e} = 0$$ as required.

Answer 4

I'd rather against the law and use l'hospital rule instead