calculate the limit $\lim_{x\rightarrow \infty}(1-\frac{1}{4x^{2}})^{x}$

Question

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: $$\lim_{x\rightarrow \infty}(1-\frac{1}{4x^{2}})^{x}=\lim_{x\rightarrow \infty}(1+\frac{-\frac{1}{4}}{x^{2}})^{x}=\sqrt[x]{\lim_{x\rightarrow \infty}(1+\frac{-\frac{1}{4}}{x^{2}})^{x})^{x}}$$ $$=\sqrt[x]{\lim_{x\rightarrow \infty}(1+\frac{-\frac{1}{4}}{x^{2}})^{x^{2}})}=\lim_{x\rightarrow \infty}\sqrt[x]{e^{\frac{-1}{4}}}=\lim_{x\rightarrow \infty}\sqrt[x]{\frac{1}{\sqrt[4]{e}}}=\lim_{x\rightarrow \infty}(\frac{1}{\sqrt[4]{e}})^{\frac{1}{x}}=1$$

Answer 1

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}

Answer 2

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 $1$).

Answer 3

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

Answer 4

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.