Evaluate the following limit using Taylor Evaluate the following limit:
\begin{equation*}
\lim_{x\to 0} \frac{e^{x^2} + 2\cos x -3}{\sin^2 x^2}.
\end{equation*}
I know the Taylor series of $e^x$ at $a=0$ is $\sum_{k=0}^{\infty} \frac{x^k}{k!}$. And if we substitute $x$ with $x^2$ we get $e^{x^2}=\sum_{k=0}^{\infty} \frac{x^{2k}}{k!}$. Also $\cos x=\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!}x^{2k}$ I am struggeling with finding the Taylor series for $\sin^2x^2$. So far I have this:
\begin{align*}
\sin^2 x=\frac{1}{2}-\frac{1}{2}\cos 2x.
\end{align*}
And by substituting $x$ with $2x$ in the Taylor series of $\cos x$ we get $\cos 2x=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}(2x)^{2k}$.
This gives us
\begin{align*}
\sin^2 x&=\frac{1}{2}-\frac{1}{2}\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}(2x)^{2k}\\
&=\frac{1}{2}-\frac{1}{2}\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}2^{2k}x^{2k}.
\end{align*}
And then, by substituting $x$ with $x^2$ we get
\begin{align*}
\sin^2 x^2 =\frac{1}{2}-\frac{1}{2}\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!}2^{2k}x^{4k}.
\end{align*}
Is this correct?
This would give:
\begin{align*}
&\lim_{x\to 0}\frac{e^{x^2} + 2\cos x -3}{\sin^2 x^2}\\
&=\lim_{x\to 0}\frac{1+x^2+O(|x|^4)+2-\frac{2x^2}{2!}+O(|x|^4)-3}{\frac{1}{2}-\frac{1}{2}(1-\frac{4x^4}{3!}+O(|x|^8))}\\
&=\lim_{x\to 0}\frac{O(|x|^4)}{\frac{2x^4}{3!}-O(|x|^8)}\\
&=\lim_{x\to 0}\frac{1}{\frac{2}{6}-O(|x|^4)}\\
&=\frac{1}{\frac{1}{3}}\\
&=3
\end{align*}
Could someone tell me if this is correct? Thanks!
 A: Neither the numerator nor the denominator has been correctly analysed.
The numerator is $O(x^4)$, but is asymptotic not to $x^4$ as your calculation incorrectly assumes, but $\tfrac12x^4+\tfrac{1}{12}x^4=\tfrac{7}{12}x^4$.
The denominator $\sin^2x^2=\tfrac12(1-\cos2x^2)$ has leading order term $\tfrac14(2x^2)^2=x^4$. You seem to have used the Taylor series for $\sin y$, rather than $\cos y$, with $y=2x^2$. An easier approach is $\sin^2x^2\sim(x^2)^2=x^4$.
So the correct limit is $\tfrac{7}{12}$, as @Atmos said.
A: Note that $\sin x\le 1$ and $\cos \ge -1.$
$$\frac{e^{x^2} + 2\cos x -3}{\sin^2 x^2}\ge{e^{x^2}-5}.$$
So
$$\lim_{x\to\infty}\frac{e^{x^2} + 2\cos x -3}{\sin^2 x^2}\ge \lim_{x\to\infty}{e^{x^2}-5}.\to\infty$$
A: Use $\sin x \approx x$ in the denominator, it becomes $x^4$, next expand numerator complete up to $x^4$.Then use $e^{z}=1+z+z^2/2+O(z^3)$, $\cos z =1-z^2/2+z^4/24+O(z^6)$ when $|z|$ is very small
$$L=\lim_{x\to\infty} \frac{e^{x^2} + 2\cos x -3}{\sin^2 x^2}=\lim_{x\to 0} \frac{1+x^2+x^4/2+2(1-x^2/2+x^4/(24)-3}{x^4}$$.
$$L=\lim_{x \to 0} \frac{7x^4/12}{x^4}=\frac{7}{12}.$$
A: $$L=\lim_{x\to 0} \frac{e^{x^2} + 2\cos x -3}{\sin^2 x^2}$$
Lets try using our dear L'Hopital's rule to simplify the expression.
$$L=\lim_{x\to 0} \frac{2xe^{x^2}  -2\sin x }{2x\sin (2x^2)}$$
We are very familiar with these functions! So lets expand them since they're easier to deal with now.
$$L=\lim_{x\to 0} \frac{x(1+x^2+\cdots)  -(x-\frac{x^3}{6}+\cdots) }{x(2x^2-\frac{8x^6}{6})}=\lim_{x\to 0} \frac{\frac{7x^3}{6}+\cdots}{2x^3+\cdots}=\bbox[10px,#ffb]{\frac{7}{12}}$$
