Find the limit : $\lim _{x \to 0}x^{\sqrt{x}}=?$

Question

Find the limit : $$\lim _{x \to 0}x^{\sqrt{x}}=?$$

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My Try :

$$f(x)=x^{\sqrt{x}}$$ $$\ln f(x)= \sqrt{x}\ln x$$

Now what ?

Answer 1

Convert to exponential

$$e^{\sqrt{x}\log(x)}$$

And you're done, since $$\sqrt{x}\log(x) \to 0$$ for $x\to 0$

Generally

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

For $n >0$

Answer 2

You have added the tag limit-without-lhospital, therefore my answer is according to that.

Let $\sqrt{x}=u$. Thus your limit $$\lim_{x \to 0} \sqrt x \ln x= \lim_{u \to 0} u \ln u^2= 2 \lim_{u \to 0} u \ln u $$

Thus, all you need to find is $\displaystyle \lim_{u \to 0^+}u\ln(u)$.

Let $u=e^{-t}$ and note that as $u \to 0^+$, we have $t \to \infty$. Hence, $$L = \lim_{u \to 0} u \ln(u) = \lim_{t \to \infty} -te^{-t} = -\lim_{t \to \infty} \dfrac{t}{e^t}$$ Now recall that $e^t \geq \dfrac{t^2}2$. Hence, we have $$\lim_{t \to \infty} \dfrac{t}{e^t} \leq \lim_{t \to \infty} \dfrac 2t = 0$$

Thus you get $$\color{red}{\lim_{x \to 0} \sqrt x \ln x= 0}$$

Answer 3

Use L'Hopital:

$$\lim_{x\to0} \sqrt{x}\ln x = \lim_{x\to0} \frac{\ln x}{\frac{1}{\sqrt{x}}} = \lim_{x\to0} \frac{\frac1x}{-\frac1{2x^{3/2}}} = -2\lim_{x\to 0} \sqrt{x} = 0$$

Now you have

$$\lim_{x\to 0} x^{\sqrt{x}} = \lim_{x\to 0} e^{\sqrt{x}\ln x} = e^0 = 1$$

using the continuity of exponentiation.

Answer 4

Note that it's in the form $$0*\infty$$ Write $\sqrt x$ as $$\frac{1}{1/\sqrt x}$$

So you get the limit of $\frac{\ln x}{1/\sqrt{x}}$ as x goes to zero. Note that it's in the form $\frac{\infty}{\infty}$. Now you can just use L'Hopital's Rule.

Answer 5

If you know that $\lim_{x\to0}x^x=1$, then just substitute $t=\sqrt{x}$, so you get $$ \lim_{t\to0}(t^2)^t= \lim_{t\to0}(t^t)^2=1^2=1 $$ For proving that $\lim_{x\to0}x\log x=0$, from which it follows that $x^x=e^{x\log x}$ has limit $e^0=1$, substitute $x=e^{-t}$, so you get $$ \lim_{t\to\infty}-\frac{t}{e^t} $$ and use the bound $e^t>1+t+t^2/2$ (for $t>0$) that you can prove with the mean value theorem, so $$ 0<\frac{t}{e^t}<\frac{t}{1+t+t^2/2} $$ and the squeeze theorem gives you the limit is $0$.

Answer 6

$\exp(\sqrt{x}\log x) =$

$ \exp(\sqrt{x}\log (\sqrt{x})^2)=$

$ \exp(2\sqrt{x}\log(\sqrt{x})).$

Let $y:=\sqrt{x}.$

Find $\lim_{y \rightarrow 0} (2y\log y) $,

and use continuity of exponential function.

Answer 7

Let's start from the basics. Using any definition of $\log x$ it is easy to show that $$\log x\leq x-1,\,x>0\tag{1}$$ If you have trouble proving the above then you need to revisit your preferred definition of $\log x$.

To evaluate your limit we can put $x=1/t$ so that $t\to\infty$ and then $$\lim_{x\to 0^{+}}\sqrt{x}\log x=-\lim_{t\to\infty} \frac{\log t} {\sqrt{t}} \tag{2}$$ We next use inequality $(1)$ to show that the limit on right hand side of $(2)$ is $0$. Let $t>1$ then $t^{1/3}>1$ and therefore by $(1)$ we have $$0<\log t^{1/3}\leq t^{1/3}-1<t^{1/3}$$ or $$0<\log t<3t^{1/3}$$ or $$0<\frac{\log t} {\sqrt{t}} <\frac{3}{t^{1/6}}$$ Applying Squeeze Theorem when $t\to\infty $ we get $$\lim_{t\to\infty} \frac{\log t} {\sqrt{t}} =0$$ and hence the limit on the left hand side of $(2)$ is also $0$.

It now follows that the limit in question is $1$.