Compute limit with the help of MacLaurin series expansion Compute $$\lim _{x \to 0} \dfrac{e^{2x^2}-1}{x^2}$$ with the aid of a MacLaurin series expansion.
 A: Edit: The problem has been corrected. We keep the original solution, and add a solution to the corrected problem below.
Hint: We have 
$$e^{(2x)^2}=1+(2x)^2+\frac{(2x)^4}{2!}+\frac{(2x)^6}{3!}+\cdots.$$
(Just write down the Maclaurin expansion of $e^t$, and everywhere in the expansion replace $t$ by $(2x)^2$.) Thus
$$\frac{e^{(2x)^2}-1}{x^2}=4+O(x^2).$$
Answer to corrected problem: This asks for 
$$\lim_{x\to 0} \frac{e^{2x^2}-1}{x^2}.$$
We have
$$e^{2x^2}=1+2x^2+\frac{(2x^2)^2}{2!}+\frac{(2x^2)^3}{3!}+\cdots.$$
Subtract $1$, divide by $x^2$. We get $2$ plus a bunch of terms that have $x$'s in them. As $x\to 0$, these terms approach $0$, so the limit is $2$.
Another way: We used the Maclaurin expansion mechanically in the solution, because it is a nice tool that it is very important to know about. But there are simpler ways. Let $t=x^2$. Then we want to find
$$\lim_{t\to 0^+}\frac{e^{2t}-1}{t}.$$
Let $f(t)=e^{2t}$. Note that $f(0)=1$. Then by the definition of the derivative, 
$$\lim_{t\to 0} \frac{f(t)-f(0)}{t}=f'(0).$$
In our case, $f'(t)=2e^{2t}$, so $f'(0)=2$, and our limit is $2$.
A: We have 
$$e^{(2x)^2}=1+(2x)^2+\frac{(2x)^4}{2!}+\frac{(2x)^6}{3!}+\cdots.$$
(Just write down the Maclaurin expansion of $e^t$, and everywhere in the expansion replace $t$ by $(2x)^2$.) Thus
$$\frac{e^{(2x)^2}-1}{x^2}=\frac{(2x)^2+\frac{(2x)^4}{2!}+..}{x^2}\cdots.$$
Now take x^2 common and cancel . The higher power terms will become zero and we'll get the answer as 4 
4 is the answer to this limit. 
