Proof that $\lim_{n\to\infty}\left(1+\frac{x^2}{n^2}\right)^{\frac{n}{2}}=1$ without L'Hospital

Question

I proved that $$\lim_{n\to\infty}\left(1+\frac{x^2}{n^2}\right)^{\frac{n}{2}}=1$$ using L'Hospital's rule. But is there a way to prove it without L'Hospital's rule? I tried splitting it as $$\lim_{n\to\infty}n^{-n}(n^2+x^2)^{\frac{n}{2}},$$ but that didn't work because $\lim_{n\to\infty}(n^2+x^2)^{\frac{n}{2}}$ diverges.

Answer 1

This has the form $\displaystyle\lim_{n\to\infty} (1+1/n)^{n}=e$.

$$\lim_{n\to\infty}\left(1+\frac{x^2}{n^2}\right)^{\frac{n}{2}}=\lim_{n\to\infty}\left(1+\frac{x^2}{n^2}\right)^{\frac{n}{2}{\color{red} {\frac{n}{x^2}\cdot\frac{x^2}{n}} }}=\lim_{n\to\infty}\left(\left(1+\frac{x^2}{n^2}\right)^{\frac{n^2}{x^2}}\right)^{{\color{red} {\frac{x^2}{2n}} }}=e^{\displaystyle\lim_{n\to\infty} \frac{x^2}{2n}} = e^0 = 1$$

Note Since $n\to\infty$ then $1/n^2$ has the same behaivor that $1/(n^2/x^2) = x^2/n^2$.

Answer 2

METHODOLOGY $1$: Direct Application of Bernoulli's Inequality

Note that for $n>|x|$

$$1\le \left(1+\frac{x^2}{n^2}\right)^{n/2}\le \frac1{\left(1-\frac{x^2}{n^2}\right)^{n/2}}\le \frac1{1-\frac{x^2}{2n}}$$

where we used Bernoulli's inequality to arrive at the last inequality.

Now apply the squeeze theorem to find

$$\lim_{n\to \infty}\left(1+\frac{x^2}{n^2}\right)^{n/2}=1$$

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METHODOLOGY $1$: Using Estimates of the Logarithm Function

Note that we may write

$$\left(1+\frac{x^2}{n^2}\right)^{n/2}=e^{(n/2)\log\left(1+\frac{x^2}{n^2}\right)}\tag 1$$

In This Answer, I used elementary, pre-calculus tools to obtain the inequalities

$$\frac{x}{1+x}\le \log(1+x)\le x \tag2$$

Using $(2)$ in $(1)$ reveals

$$e^{nx^2/(2n^2+2x^2)}\le e^{(n/2)\log\left(1+\frac{x^2}{n^2}\right)}\le e^{x^2/2n}$$

whence application of the squeeze theorem yields the coveted result

$$\lim_{n\to \infty}\left(1+\frac{x^2}{n^2}\right)^{n/2}=1$$

as expected!

Answer 3

Consider the following for large n and finite x: $$e^{\frac{x^2}{n^2}} \approx 1+\frac{x^2}{n^2}$$ Therefore, rewrite the limit as: $$\lim_{n \to \infty} {\left(e^{\frac{x^2}{n^2}}\right)}^{\frac{n}{2}}$$ $$=\lim_{n \to \infty} e^{\frac{x^2}{2n}}$$ $$=1$$

Answer 4

I have an algebraic solution. Let's our limit be $L$: $$L=\lim_{n\rightarrow\infty}\left(1+\frac{x^2}{n^2}\right)^\frac{n}{2}$$ Now, we make two changes of variables: $$t = \frac{n}{2} $$ and $$y=\frac{x^2}{4}$$ Then we have: $$L=\lim_{n\rightarrow\infty}\left(1+\frac{x^2}{n^2}\right)^\frac{n}{2}=\lim_{t\rightarrow\infty}\left(1+\frac{x^2}{2t^2}\right)^t=\lim_{t\rightarrow\infty}\left(1+\frac{y}{t^2}\right)^t$$ Then, rewrite our limit as: $$L=\lim_{t\rightarrow\infty}\left(1+\frac{y}{t^2}\right)^t=\lim_{t\rightarrow\infty}e^{t \ln{\left(1+\frac{y}{t^2}\right)}}=e^{\lim_{t\rightarrow\infty} t \ln{\left(1+\frac{y}{t^2}\right)}}=e^{L_1}$$ Where $L_1=\lim_{t\rightarrow\infty} t \ln{\left(1+\frac{y}{t^2}\right)}$

Now, we make another change of variables: $$r=1/t^2$$ $$L_1=\lim_{t\rightarrow\infty} t \ln{\left(1+\frac{y}{t^2}\right)}=\lim_{t\rightarrow0} \frac{\ln{\left(1+ry\right)}}{\sqrt{r}}=\lim_{r\rightarrow0} \frac{\ln{\left(1+ry\right)}}{yr} \frac{yr}{\sqrt{r}}=\lim_{r\rightarrow0} y\sqrt{r}=0$$ Finally: $$L=\lim_{n\rightarrow\infty}\left(1+\frac{x^2}{n^2}\right)^\frac{n}{2}=e^{L_1}=e^0=1$$

Answer 5

For the upper bound using Bernoulli inequality note that it applies for exponents $t: t \leq 0 \cup t \geq 1$, so for $\frac{n}{2} < 0$: $$ \bigg(1+\frac{x^2}{n^2} \bigg)^\frac{n}{2}= \frac{1}{\bigg(1+\frac{x^2}{n^2} \bigg)^{-\frac{n}{2}}} \leq \frac{1}{1- \frac{x^2}{2n}} \to 1 $$ And the limit follows to squeeze lemma

Answer 6

The lemma of Thomas Andrews can be used here:

Lemma: If $n(a_n-1)\to 0$ then $a_n^n\to 1$.

Now use this with $$a_n=\sqrt{1+\frac{x^2}{n^2}}$$

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Perhaps you are trying to deal with the limit of $(1+ix/n)^n$ and show that it equals $\cos x+i\sin x$. That can also be easily handled by the lemma in question without first dealing with $|(1+ix/n)^n|$. Just apply the lemma to $$a_n=\dfrac{1 +\dfrac{ix} {n}} {\cos\dfrac{x} {n} +i\sin\dfrac{x} {n}} $$

Answer 7

This follows from the fact that for any sequence (real or complex) $z_n\rightarrow z$,

$$ \Big(1+\frac{z_n}{n}\Big)^n\xrightarrow{n\rightarrow\infty}e^z $$