Solving $\lim_{ x \to 0^+} \sqrt{\tan x}^{\sqrt{x}}$

Question

The limit to be calculated is:

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

I tried:

$$L =\lim_{x \to 0^+}\sqrt{\tan x}^{\sqrt{x}}$$

$$\log L = \lim _{x\to 0^+} \ \ \dfrac{1}{2}\cdot\dfrac{\sqrt{x}}{\frac{1}{\log(\tan x)}}$$

To apply the L'hospital theorem but failed, as it went more complex.

How can we solve this, with or without L'Hospital?

Answer 1

Well, we have $$ \begin{align}\lim_{x\to 0^+}\sqrt{x}\ln\sqrt{\tan x}&=\lim_{x\to 0^+}\frac{\sqrt{x}}{2}\ln(\tan x)\\ &=\lim_{x\to 0^+}\frac{\ln(\tan x)}{\frac{2}{\sqrt{x}}}\qquad\text{which takes the form } \frac{-\infty}{+\infty}\text{and apply LHR to get}\\ &=\lim_{x\to 0^+}\frac{\frac{\sec^2x}{\tan x}}{-\frac{1}{\sqrt{x^3}}}\\ &=\lim_{x\to 0^+}\frac{\frac{2}{\sin 2x}}{-\frac{1}{x\sqrt{x}}}\\ &=\lim_{x\to 0^+}\bigg[-\sqrt{x}\cdot\frac{2x}{\sin 2x}\bigg]\\ &=-\sqrt{0}\cdot 1=0. \end{align}$$ Hence, the required limit is $$\lim_{x \to 0^+}\sqrt{\tan x}^{\sqrt{x}}=e^{\lim_{x \to 0^+}\ln\sqrt{\tan x}^{\sqrt{x}}}=e^{\lim_{x\to 0^+}\sqrt{x}\ln\sqrt{\tan x}}=e^0=1.$$

Answer 2

HINT: write $$e^{\lim_{x\to 0}\frac{\ln(\sqrt{\tan(x)}}{\frac{1}{\sqrt{x}}}}$$ and use L'Hospital

Answer 3

$$\begin{align} \left(\sqrt{\tan(x)}\right)^{\sqrt{x}}&=e^{\frac12\sqrt{x}\log(\sin(x))-\frac12\sqrt{x}\log(\cos(x))}\\\\ &\underbrace{e^{-\frac12\sqrt{x}\log(\cos(x))}}_{\to 1}\,\underbrace{e^{\frac12\sqrt{x}\log\left(\frac{\sin(x)}{x}\right)}}_{\to 1}\,\underbrace{e^{\sqrt{x}\log(\sqrt{x})}}_{\to 1}\\\\ \end{align}$$

since we recall that $\lim_{y\to 0}y\log(y)=0$ (i.e., with $y=\sqrt{x}$), which can be shown a variety of ways including application of L'Hospital's Rule.

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Note that we could have arrived at this result quickly by using asymptotic analysis.

Here $\tan(x)\sim x$ as $x\to 0$.

Then we are evaluating equivalently $\lim_{x\to 0}(\sqrt{x})^{\sqrt{x}}$ which by substitution $x=y^2$ becomes equal to $\lim_{y\to 0}y^y=1$

Answer 4

For small $x>0,$ we can apply $\ln$ to get

$$0\ge \tag 1 \sqrt x \ln \sqrt {\tan x} = (\sqrt x)/2\cdot \ln\tan x$$

Now $\tan x > x.$ Thus $\ln\tan x > \ln x.$ It follows that

$$\tag 2 0\ge \sqrt x \ln \sqrt {\tan x} \ge \sqrt x/2 \cdot \ln x.$$

Now use the well known limit $\lim_{x\to 0^+} x^p\ln x =0$ for any $p>0$ to see that the limit of the right side of $(2)$ is $0.$ By the squeeze theorem, $\lim_{x\to 0^+}\sqrt x \ln \sqrt {\tan x} =0.$ Exponentiating back shows the limit of interest is $1.$

Answer 5

Suppose you know $u^u\to1$ as $u\to0^+$ and ${\tan\theta\over\theta}\to1$ as $\theta\to0$. Then, by writing $x=4u^2$, we have

$$\sqrt{\tan x}^{\sqrt x}=(\tan(4u^2))^u=4^u(u^u)^2\left(\tan(4u^2)\over4u^2\right)^u\to4^0(1)^2(1)^0=1\quad\text{as }u\text{ (hence also }x)\to0^+$$