# Limit, Taylor

#### Petrus

##### Well-known member
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
Re: limit,taylor

I think you mean x goes to zero right ?

#### Petrus

##### Well-known member
Re: limit,taylor

I think you mean x goes to zero right ?

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
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
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
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
Re: limit,taylor

I am making so much typo.. I did Edit it now!!
I am sure you can solve it yourself

#### Petrus

##### Well-known member
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,
$$\displaystyle |\pi\rangle$$