Limit with square and cube root

wolly

Junior Member
Joined
Jul 18, 2018
Messages
116
limx0cos(x)cos(x)3sin(x)ln(1+sin(x))\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 a3b3=(ab)(a2+ab+b2)a^3-b^3=(a-b)(a^2+ab+b^2)?
 
Can You apply this to x4\sqrt[4]{x} x5\sqrt[5]{x}...xn\sqrt[n]{x}?
 
Can You apply this to x4\sqrt[4]{x} x5\sqrt[5]{x}...xn\sqrt[n]{x}?
Your problem is in the denominator [you need to get rid of ln(1+sin(x))\ln(1 + \sin(x))], not in the numerator. Therefore, I don't think that will help you.

hint: ln(1+x)x when x0\ln(1 + x) \rightarrow x \ \text{when} \ x \rightarrow 0.
 
Well the denominator îs x^2 because
limx0sin(x)ln(1+sin(x))\lim_{x \to 0}sin(x)*ln(1+sin(x))=
limx0sin(x)xxln(1+sin(x))sin(x)sin(x)xx=x2\lim_{x \to 0}\frac{sin(x)}{x}*x*\frac{ln(1+sin(x))}{sin(x)}*\frac{sin(x)}{x}*x=x^{2}
 
Last edited:
Well the denominator îs x^2 because
limx0sin(x)ln(1+sin(x))\lim_{x \to 0}sin(x)*ln(1+sin(x))=
sin(x)xxln(1+sin(x)sin(x)sin(x)xx=x2\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:

limx0cos(x)cos(x)3sin(x)ln(1+sin(x))=limx0cos(x)cos(x)3x2\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 cos(x) and cos(x)3\displaystyle \sqrt{\cos(x)} \ \text{and} \ \sqrt[3]{\cos(x)} in terms of xx.

Hint: Maclaurin series.
 
Correct.

Now rewrite your limit with this simplification:

limx0cos(x)cos(x)3sin(x)ln(1+sin(x))=limx0cos(x)cos(x)3x2\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
a3b3=(ab)(a2+ab+b2)a^3-b^3=(a-b)(a^2+ab+b^2) and
a3+b3=(a+b)(a2ab+b2)a^3+b^3=(a+b)(a^2-ab+b^2)?
The calculator used both of them.
 
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
a3b3=(ab)(a2+ab+b2)a^3-b^3=(a-b)(a^2+ab+b^2) and
a3+b3=(a+b)(a2ab+b2)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?
 
So în my case a=cos(x)a=\sqrt{cos(x)} and b=cos(x)3b=\sqrt[3]{cos(x)}?
 
I mean I tried to solve it and the answer wasn't the same with cos2x(cos(x)1)x2\frac{cos^{2}x(cos(x)-1)}{x^2}
 
So în my case a=cos(x)a=\sqrt{cos(x)} and b=cos(x)3b=\sqrt[3]{cos(x)}?
I don't see how that will work, but I have an idea.

ln(1+sinx)sinx when x0\ln(1 + \sin x) \rightarrow \sin x \ \text{when} \ x \rightarrow 0


limx0cosxcosx3sinxln(1+sinx)=limx0cosxcosx3sin2x=limx0cosxcosx31cos2x\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 t2=cosxt^2 = \cos x


limx0cosxcosx31cos2x=limt1tt2/31t4=limt1t2/3(t1/31)(1t)(1+t+t2+t3)\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)}


t1/3113(t1) as t1t^{1/3} - 1 \approx \frac{1}{3}(t - 1) \ \text{as} \ t\rightarrow 1


limt1t2/3(t1/31)(1t)(1+t+t2+t3)=limt113t2/3(t1)(1t)(1+t+t2+t3)=limt113t2/3(1+t+t2+t3)=13(1)(1+1+1+1)=112\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}
 
Top