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Limit, Taylor

Petrus

Well-known member
Feb 21, 2013
739
Hello MHB,
I am currently working with a old exam and it says.


\(\displaystyle \lim_{x->0}\frac{3-e^{x^2}-2\cos(x)}{x^2\sin^2(x)}\)
and this is how far I got but struggle on the simplify

Regards,
\(\displaystyle |\pi\rangle\)
 
Last edited:

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: limit,taylor

I think you mean x goes to zero right ?
 

Petrus

Well-known member
Feb 21, 2013
739

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: limit,taylor

\(\displaystyle \lim_{x \to 0}\frac{(3-1-x^2-\frac{x^4}{2!}-\cdot \cdot \cdot ) -2+x^2-2 \cdot \frac{x^4}{4!} +\cdot \cdot \cdot }{x^3-\frac{x^5}{3!}+\cdot \cdot \cdot}\)

Sounds like the numerator dominates, unless you meant sin^2(x) in the denominator ..
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: limit,taylor

By the way , in the case of a limit we have to use the expansion of all functions near the point , which is approached by x , hence the functions must converge .
 

Petrus

Well-known member
Feb 21, 2013
739
Re: limit,taylor

\(\displaystyle \lim_{x \to 0}\frac{(3-1-x^2-\frac{x^4}{2!}-\cdot \cdot \cdot ) -2+x^2-2 \cdot \frac{x^4}{4!} +\cdot \cdot \cdot }{x^3-\frac{x^5}{3!}+\cdot \cdot \cdot}\)

Sounds like the numerator dominates, unless you meant sin^2(x) in the denominator ..
I am making so much typo..:( I did Edit it now!!
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667

Petrus

Well-known member
Feb 21, 2013
739
Re: limit,taylor

I am sure you can solve it yourself :)
Actually I believe you are correct! I did accidently had a sign wrong.. If someone is intrested how to solve it. You can first cancel out all \(\displaystyle x^0, 3-1-2\) then you can also cancel out \(\displaystyle x^2, -x^2+\frac{2x^2}{2!}\) expand that square function at numerator and divide evrything by \(\displaystyle x^4\) and you get the answer \(\displaystyle - \frac{7}{12}\), what I did wrong is that I did not see the negative sign in bracket at top. I did take \(\displaystyle \frac{1}{2}-\frac{2}{4!}\) while it should be \(\displaystyle -\frac{1}{2}-\frac{2}{4!}\)
Thanks once again Zaid!:)

Regards,
\(\displaystyle |\pi\rangle\)