How do you evaluate this limit?

[Leo Liu]

Homework Statement

$$\lim_{x\to 0^+}(1-\cos (\sqrt x))^{\sin(x)}$$

Relevant Equations

.

I tried taking e^ln but to no avail. Please help! Thanks.

My attempt:
$$\lim_{x\to 0^+}(1-\cos (\sqrt x))^{\sin(x)}$$
$$\lim_{x\to 0^+}e^{\ln (1-\cos\sqrt x)^{\sin x}}$$
$$\lim_{x\to 0^+}e^{{\sin x}\ln (1-\cos\sqrt x)}$$
$$\lim_{x\to 0^+}\exp(\frac{\ln (1-\cos\sqrt x)}{1/\sin x})$$
If I apply Lhospital's rule to this limit, the result will be quite complicated and will remain in intermediate form. I also reached wolfram alpha for help, yet the step by step solution is terse.

 

[PeroK]

Homework Statement:: $$\lim_{x\to 0^+}(1-\cos (\sqrt x))^{\sin(x)}$$
Relevant Equations:: .

I tried taking e^ln but to no avail. Please help! Thanks.

My attempt:
$$\lim_{x\to 0^+}(1-\cos (\sqrt x))^{\sin(x)}$$
$$\lim_{x\to 0^+}e^{\ln (1-\cos\sqrt x)^{\sin x}}$$
$$\lim_{x\to 0^+}e^{{\sin x}\ln (1-\cos\sqrt x)}$$
$$\lim_{x\to 0^+}\exp(\frac{\ln (1-\cos\sqrt x)}{1/\sin x})$$
If I apply Lhospital's rule to this limit, the result will be quite complicated and will remain in intermediate form. I also reached wolfram alpha for help, yet the step by step solution is terse.

Just keep going with l'Hopital, complicated or not.

You don't need the exponential.

 

[vela]

You could use the trig identity $\sin^2 \frac \theta 2 = \frac{1-\cos \theta}{2}$ to simplify the numerator before differentiating.

 

[Leo Liu]

Just keep going with l'Hopital, complicated or not.

You don't need the exponential.

I am lazy so I let MMA do the job for me. Here is what I got after differentiating the fraction twice:

Although the graph verifies that the limit is indeed 0 as x approaches 0 from right side, it seems that Num2/Den2 is still in intermediate form. I doubt that we will obtain an answer by blindly differentiating the fraction. Thanks anyway.

 

[Office_Shredder]

Do you know what the Taylor series for sine and cosine are, or have you not reached that part in your class yet?

 

[vela]

After the first round of differentiation, you should be able to show that you end up with
$$-\frac 14 \cdot \frac{\sin \sqrt x}{\sqrt x} \cdot \frac{\sin x}{\sin \frac{\sqrt x}2} \cdot \frac{\tan x}{\sin \frac{\sqrt x}2}.$$ Calculate the limit of each factor.

 

[Leo Liu]

After the first round of differentiation, you should be able to show that you end up with
$$-\frac 14 \cdot \frac{\sin \sqrt x}{\sqrt x} \cdot \frac{\sin x}{\sin \frac{\sqrt x}2} \cdot \frac{\tan x}{\sin \frac{\sqrt x}2}.$$ Calculate the limit of each factor.

This is helpful. Thank you. I didn't expect this question could be so complex.

 

[anuttarasammyak]

By Taylor expansion it is
\lim_{x \rightarrow +0} (\frac{x}{2}+o(x^2))^{x+o(x^3)}=\lim_{x \rightarrow +0} x^x=1-0

 

[Leo Liu]

Someone on reddit sent me his work:

So the key is to use the substitution $x=t^2$ to simplify the expression and to notice $\lim_{t\to 0} 2t/\sin t=2$ before applying the product rule of limit.
Credit to @vladislavsrb