Limit with square and cube root

[wolly]

$$\lim_{x \to 0}\frac{\sqrt{cos(x)}-\sqrt[3]{cos(x)}}{sin(x)ln(1+sin(x))}$$
Can I conjugate the square Root and cube Root using the formula $$a^3-b^3=(a-b)(a^2+ab+b^2)$$?

 

[wolly]

Can You apply this to $$\sqrt[4]{x}$$ $$\sqrt[5]{x}$$...$$\sqrt[n]{x}$$?

 

[mario99]

Can You apply this to $$\sqrt[4]{x}$$ $$\sqrt[5]{x}$$...$$\sqrt[n]{x}$$?

Your problem is in the denominator [you need to get rid of $\ln(1 + \sin(x))$], not in the numerator. Therefore, I don't think that will help you.

hint: $\ln(1 + x) \rightarrow x \ \text{when} \ x \rightarrow 0$.

 

[wolly]

Your problem is in the denominator [you need to get rid of $\ln(1 + \sin(x))$], not in the numerator. Therefore, I don't think that will help you.

hint: $\ln(1 + x) \rightarrow x \ \text{when} \ x \rightarrow 0$.

Well the denominator îs x^2 because
$$\lim_{x \to 0}sin(x)*ln(1+sin(x))$$=
$$\lim_{x \to 0}\frac{sin(x)}{x}*x*\frac{ln(1+sin(x))}{sin(x)}*\frac{sin(x)}{x}*x=x^{2}$$

 

[wolly]

You didn't answered my question @mario99

 

[mario99]

Well the denominator îs x^2 because
$$\lim_{x \to 0}sin(x)*ln(1+sin(x))$$=
$$\frac{sin(x)}{x}*x*\frac{ln(1+sin(x)}{sin(x)}*\frac{sin(x)}{x}*x=x^{2}$$

Correct.

Now rewrite your limit with this simplification:

$\displaystyle \lim_{x \rightarrow 0}\frac{\sqrt{\cos(x)}-\sqrt[3]{\cos(x)}}{\sin(x)\ln(1+\sin(x))} = \lim_{x \rightarrow 0}\frac{\sqrt{\cos(x)}-\sqrt[3]{\cos(x)}}{x^2}$


Now you need to find a way to write $\displaystyle \sqrt{\cos(x)} \ \text{and} \ \sqrt[3]{\cos(x)}$ in terms of $x$.

Hint: Maclaurin series.

 

[mario99]

You didn't answered my question @mario99

Can I conjugate the square Root and cube Root using the formula $$a^3-b^3=(a-b)(a^2+ab+b^2)$$?

If you mean this by your question. I don't see how it helps. If you think that it helps, try it.

 

[wolly]

Correct.

Now rewrite your limit with this simplification:

$\displaystyle \lim_{x \rightarrow 0}\frac{\sqrt{\cos(x)}-\sqrt[3]{\cos(x)}}{\sin(x)\ln(1+\sin(x))} = \lim_{x \rightarrow 0}\frac{\sqrt{\cos(x)}-\sqrt[3]{\cos(x)}}{x^2}$

Conjugation?
I used a limit calculator because they didn't teached us at highschool conjugation with hingher orders.
I'm clueless and we can't use differentiation because they didn't teached us.
Do You use both
$$a^3-b^3=(a-b)(a^2+ab+b^2)$$ and
$$a^3+b^3=(a+b)(a^2-ab+b^2)$$?
The calculator used both of them.

 

[mario99]

Conjugation?
I used a limit calculator because they didn't teached us at highschool conjugation with hingher orders.
I'm clueless and we can't use differentiation because they didn't teached us.
Do You use both
$$a^3-b^3=(a-b)(a^2+ab+b^2)$$ and
$$a^3+b^3=(a+b)(a^2-ab+b^2)$$?
The calculator used both of them.

If the calculator has used conjugation, there must be a way to solve the problem by them, but I cannot see it. You are not allowed to use infinite series?

 

[wolly]

So în my case $$a=\sqrt{cos(x)}$$ and $$b=\sqrt[3]{cos(x)}$$?

 

[wolly]

I mean I tried to solve it and the answer wasn't the same with $$\frac{cos^{2}x(cos(x)-1)}{x^2}$$

 

[mario99]

So în my case $$a=\sqrt{cos(x)}$$ and $$b=\sqrt[3]{cos(x)}$$?

I don't see how that will work, but I have an idea.

$\ln(1 + \sin x) \rightarrow \sin x \ \text{when} \ x \rightarrow 0$


$\displaystyle \lim_{x\rightarrow 0}\frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin x \ln(1 + \sin x)} = \lim_{x\rightarrow 0}\frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2 x} = \lim_{x\rightarrow 0}\frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{1 - \cos^2 x}$


Let $t^2 = \cos x$


$\displaystyle \lim_{x\rightarrow 0}\frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{1 - \cos^2 x} = \lim_{t\rightarrow 1}\frac{t - t^{2/3}}{1 - t^4} = \lim_{t\rightarrow 1}\frac{t^{2/3}(t^{1/3} - 1)}{(1 - t)(1 + t + t^2 + t^3)}$


$t^{1/3} - 1 \approx \frac{1}{3}(t - 1) \ \text{as} \ t\rightarrow 1$


$\displaystyle \lim_{t\rightarrow 1}\frac{t^{2/3}(t^{1/3} - 1)}{(1 - t)(1 + t + t^2 + t^3)} = \lim_{t\rightarrow 1}\frac{\frac{1}{3}t^{2/3}(t - 1)}{(1 - t)(1 + t + t^2 + t^3)} = \lim_{t\rightarrow 1}\frac{\frac{1}{3}t^{2/3}}{-(1 + t + t^2 + t^3)} = \frac{\frac{1}{3}(1)}{-(1 + 1 + 1 + 1)} = -\frac{1}{12}$