Limit of function

[Anakin_99]

Hi guys I can' t understand where I am wrong in this limit, thank u if u can help me, answer should be 1/6

 

[Dr.Peterson]

Hi guys I can' t understand where I am wrong in this limit, thank u if u can help me, answer should be 1/6

View attachment 28023

Pretend you are one of us, reading through this and looking for an error. At each line, ask what was being done.

To start, what is the Taylor series for the sine?

 

[lex]

This only takes two lines:
[MATH]\sin x^3=x^3-\frac{x^9}{6}+O(x^{15})\hspace2ex[/MATH](substituting [MATH]x^3[/MATH] for [MATH]x[/MATH] in the expansion of [MATH]\sin x[/MATH])
...

 

[pka]

The answer is indeed \(\dfrac{1}{6}\) to get it you need to apply L'Hopital's rule at least five times.

 

[Anakin_99]

i just noticed I used the cos development instead of the sin one but still cant end the exercise..

 

[lex]

You know that [MATH]\sin X=X-\frac{X^3}{6}+O(X^5)[/MATH]so substituting [MATH]X=x^3: \hspace3ex \sin (x^3)=(x^3)-\frac{(x^3)^3}{6}+O{(x^3)^5} \hspace3ex[/MATH](remembering [MATH](x^3)^3=x^{3\times3}=x^9[/MATH])
[MATH]\sin x^3=x^3-\frac{x^9}{6} + O(x^{15})[/MATH]

 

[Dr.Peterson]

You made a second mistake in your first attempt, besides using the wrong series; and you made the same error again here! \(\left(x^3\right)^2 \ne x^{(3^2)}\), and \(\left(x^3\right)^3 \ne x^{(3^3)}\)!

 

[lookagain]

Anakin_99, if you are not instructed to use Taylor's series and are allowed to use
L'Hopital's Rule, consider:

Let \(\displaystyle \ u = x^3.\)

Use L'Hopital's Rule twice, and then you'll have the limit as u ---> 0+ of

\(\displaystyle \dfrac{sin(u)}{6u} \ \ \) to evaluate to finish it.