calculate the limit $\lim_{x\rightarrow \infty}(1-\frac{1}{4x^{2}})^{x}$ Just to be sure if what i have done is correct if i have done something wrong please tell me so.
I have also used the fact that : $\lim_{x\rightarrow  \infty}(1+\frac{a}{x})^{x}=e^{a}$
solution:
$$\lim_{x\rightarrow  \infty}(1-\frac{1}{4x^{2}})^{x}=\lim_{x\rightarrow  \infty}(1+\frac{-\frac{1}{4}}{x^{2}})^{x}=\sqrt[x]{\lim_{x\rightarrow  \infty}(1+\frac{-\frac{1}{4}}{x^{2}})^{x})^{x}}$$
$$=\sqrt[x]{\lim_{x\rightarrow  \infty}(1+\frac{-\frac{1}{4}}{x^{2}})^{x^{2}})}=\lim_{x\rightarrow  \infty}\sqrt[x]{e^{\frac{-1}{4}}}=\lim_{x\rightarrow  \infty}\sqrt[x]{\frac{1}{\sqrt[4]{e}}}=\lim_{x\rightarrow  \infty}(\frac{1}{\sqrt[4]{e}})^{\frac{1}{x}}=1$$
 A: Your solution definitely works, but it's just that you've chosen a rather circuitous way to do it. A more straightforward way to find this limit would be to notice that the expression $1-\frac{1}{4x^{2}}$ is nothing but a difference of squares. Armed with that observation, here's how you solve it:
\begin{align}
\lim_{x\rightarrow\infty} \left(1-\frac{1}{4x^{2}}\right)^{x}
&=\lim_{x\rightarrow\infty} \left[1-\left(\frac{1}{2x}\right)^2\right]^{x}\\
&=\lim_{x\rightarrow\infty} \left[\left(1-\frac{1}{2x}\right)\left(1+\frac{1}{2x}\right)\right]^{x}\\
&=\lim_{x\rightarrow\infty} \left[\left(1-\frac{1}{2x}\right)^{x}\left(1+\frac{1}{2x}\right)^{x}\right]\\
&=\lim_{x\rightarrow\infty} \left(1+\frac{-1/2}{x}\right)^x\cdot
\lim_{x\rightarrow\infty} \left(1+\frac{1/2}{x}\right)^x\\
&=e^{-\frac{1}{2}}\cdot e^{\frac{1}{2}}\\
&=e^{-\frac{1}{2} + \frac{1}{2}}\\
&=e^{0}\\
&=1\\
\end{align}
A: Hint
Don't use the notation $\sqrt[x]{a}$ for $x\notin \mathbb N$.
The second equality is wrong a priori.
$$\lim_{x\to \infty }\left(1-\frac{1}{4x^2}\right)^x=\lim_{x\to \infty }e^{x\ln\left(1-\frac{1}{4x^2}\right)}.$$
Now, using $$\lim_{u\to 0 }\frac{\ln(1-u^2)}{u^2}=-1,$$
and the fact that $x\longmapsto e^x$ is continuous at $0$, the claim follow (and the final result is indeed $1$).
A: The simpler results for the limits of the form $1^\infty$ will be
$$\lim_{x\to \infty }{(1+f(x))^{g(x)}} $$ where $f(x) \rightarrow0$ and $g(x)\rightarrow \infty$ would be
$$\lim_{x\to \infty }e^{g(x)(f(x)-1)}$$
So, using this technique, your answer i.e., $1$ is right
A: Here is a smart way to avoid $e$ altogether. First note that if $n$ is a positive integer then Bernoulli's inequality gives us $$1-\frac{1}{4n}\leq \left(1-\frac{1}{4n^{2}}\right)^{n}\leq 1$$ and therefore by Squeeze Theorem the limit in question $1$ if $x$ is an integer variable. For real variable $x$ we use the inequalities $$\left(1-\frac{1}{4[x]^{2}}\right)^{[x]+1}\leq \left(1-\frac{1}{4x^{2}}\right)^{x}\leq\left(1-\frac{1}{4([x]+1)^{2}}\right)^{[x]}$$ and by Squeeze Theorem we are done. 
