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Limit with square and cube root
- Thread starter wolly
- Start date
Your problem is in the denominator [you need to get rid of [imath]\ln(1 + \sin(x))[/imath]], not in the numerator. Therefore, I don't think that will help you.Can You apply this to [math]\sqrt[4]{x}[/math] [math]\sqrt[5]{x}[/math]...[math]\sqrt[n]{x}[/math]?
hint: [imath]\ln(1 + x) \rightarrow x \ \text{when} \ x \rightarrow 0[/imath].
Well the denominator îs x^2 becauseYour problem is in the denominator [you need to get rid of [imath]\ln(1 + \sin(x))[/imath]], not in the numerator. Therefore, I don't think that will help you.
hint: [imath]\ln(1 + x) \rightarrow x \ \text{when} \ x \rightarrow 0[/imath].
[math]\lim_{x \to 0}sin(x)*ln(1+sin(x))[/math]=
[math]\lim_{x \to 0}\frac{sin(x)}{x}*x*\frac{ln(1+sin(x))}{sin(x)}*\frac{sin(x)}{x}*x=x^{2}[/math]
Last edited:
Correct.Well the denominator îs x^2 because
[math]\lim_{x \to 0}sin(x)*ln(1+sin(x))[/math]=
[math]\frac{sin(x)}{x}*x*\frac{ln(1+sin(x)}{sin(x)}*\frac{sin(x)}{x}*x=x^{2}[/math]
Now rewrite your limit with this simplification:
[imath]\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}[/imath]
Now you need to find a way to write [imath]\displaystyle \sqrt{\cos(x)} \ \text{and} \ \sqrt[3]{\cos(x)}[/imath] in terms of [imath]x[/imath].
Hint: Maclaurin series.
You didn't answered my question @mario99
If you mean this by your question. I don't see how it helps. If you think that it helps, try it.Can I conjugate the square Root and cube Root using the formula [math]a^3-b^3=(a-b)(a^2+ab+b^2)[/math]?
Conjugation?Correct.
Now rewrite your limit with this simplification:
[imath]\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}[/imath]
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
[math]a^3-b^3=(a-b)(a^2+ab+b^2)[/math] and
[math]a^3+b^3=(a+b)(a^2-ab+b^2)[/math]?
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?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
[math]a^3-b^3=(a-b)(a^2+ab+b^2)[/math] and
[math]a^3+b^3=(a+b)(a^2-ab+b^2)[/math]?
The calculator used both of them.
I don't see how that will work, but I have an idea.So în my case [math]a=\sqrt{cos(x)}[/math] and [math]b=\sqrt[3]{cos(x)}[/math]?
[imath]\ln(1 + \sin x) \rightarrow \sin x \ \text{when} \ x \rightarrow 0[/imath]
[imath]\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}[/imath]
Let [imath]t^2 = \cos x[/imath]
[imath]\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)}[/imath]
[imath]t^{1/3} - 1 \approx \frac{1}{3}(t - 1) \ \text{as} \ t\rightarrow 1[/imath]
[imath]\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}[/imath]