Thread: Taylor Series on a limit

1. Taylor Series on a limit

I'm generally ok with these sorts of problems, but these two have me stumped. Our professor hasn't gone over these particular cases (taylor series on a limit/integral) so any help is appreciated.

Here I have to use Taylor series to evaluate:

L'hopital doesnt get me anywhere on this limit. My Ti-89 says the limit is equal to 3 but that doesnt make sense. All the derivatives f'(3), f''(0) etc.. would just be 0 after making Taylors series pointless.

and,

Find the first four terms for the MacLaurin (Taylor series evaluated around x = 0) series for

Sumner

2. In the first one, you're supposed to use the Taylor series for $e^{x^2}$ and $\sin{x}$; namely, as $e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots$, then $e^{x^2}=1+x^2+\frac{x^4}{2}+\frac{x^6}{6}+\cdots$; and as $\sin{x}=x-\frac{x^3}{6}+\frac{x^5}{120}-\cdots$, then $c=x^2-\frac{x^4}{6}+\frac{x^6}{120}+\cdots$.

Thus $\lim_{x\to0}\frac{e^{x^2}-1-x^2}{x\sin{x}-x^2}=\lim_{x\to0}\frac{\frac{x^4}{2}+\frac{x^6}{6} +\cdots}{-\frac{x^4}{6}+\frac{x^6}{120}+\cdots}=\lim_{x\to0} \frac{\frac{1}{2}+\frac{x^2}{6}+\cdots}{-\frac{1}{6}+\frac{x^2}{120}+\cdots}=-\frac{6}{2}=-3$.

--Kevin C.

3. Originally Posted by white-tt
I'm generally ok with these sorts of problems, but these two have me stumped. Our professor hasn't gone over these particular cases (taylor series on a limit/integral) so any help is appreciated.

Here I have to use Taylor series to evaluate:

L'hopital doesnt get me anywhere on this limit. My Ti-89 says the limit is equal to 3 but that doesnt make sense. All the derivatives f'(3), f''(0) etc.. would just be 0 after making Taylors series pointless.
follow the instructions. after writing out the Taylor expansions for $e^{x^2}$ and $\sin x$, there will be no need of L'Hopital's rule. it will simply be a ratio of polynomials, which you should be able to deal with

and,

Find the first four terms for the MacLaurin (Taylor series evaluated around x = 0) series for

Sumner
do you remember how to find the coefficients of the Taylor series? Hint: for the first derivative, use the second fundamental theorem of calculus

4. In the second, remember the definition of the Maclaurin series for f(x):
$f(0)+f'(0)x+\frac{f''(0)}{2}x^2+\frac{f'''(0)}{6}x ^2+\cdots$. Now, if $f(x)=\int_{0}^{x}\sqrt{1+t^3}\,dx$, then we see $f(0)=\int_{0}^{0}\sqrt{1+t^3}\,dx=0$. The Fundamental theorem of calculus tells us that $f'(x)=\sqrt{1+x^3}$, and from there you can find the higher derivatives, and thus the Maclaurin series.

--Kevin C.

5. A good alternative for arriving to the Mc Laurin expansion of...

$f(x)= \int_{0}^{x} \sqrt{1+t^{3}}\cdot dt$ (1)

$\sqrt{1+a}= 1 + \frac{1}{2}\cdot a - \frac{1}{2\cdot 4}\cdot a^{2} + \frac{1\cdot 3}{2\cdot 4\cdot 6}\cdot a^{3} - \dots$ (2)

... then you set in (2) $a=t^{3}$ and finally integrate 'term by term' ...

Kind regards

$\chi$ $\sigma$

6. Wow! Thanks for your help everybody, I didnt think of using expansions on the first one, that makes sense now.

You guys are a lifesaver!

Thanks again!!

Sumner

7. Hi

These 2 problems have already been posted recently

http://www.mathhelpforum.com/math-he...or-series.html

http://www.mathhelpforum.com/math-he...-integral.html

I guess generic1 and you are in the same class !

8. Originally Posted by running-gag
Hi

These 2 problems have already been posted recently

http://www.mathhelpforum.com/math-he...or-series.html

http://www.mathhelpforum.com/math-he...-integral.html

I guess generic1 and you are in the same class !

haha doesnt surprise me! Everybody's been pretty lost on these. Sorry for the double post then, guys! I tried searching but didnt come across those.