Evaluating $\lim_{x\rightarrow \pm \infty} \Big(1\pm \frac a{x}\Big)^x$, with $a>0$, for each choice of sign We know that:
$$e=\lim_{x\rightarrow \infty} \Big(1+\frac 1{x}\Big)^x$$
But I have some problems with these, with independent choices of sign:
$$\lim_{x\rightarrow \pm \infty} \Big(1\pm \frac a{x}\Big)^x \ \text{for } a\gt 0$$
I will show how I evaluate each one.
Case 1 ($+\infty$, $+a/x$):
$$
\lim_{x\rightarrow \infty}\Big(1+\frac a{x}\Big)^x \\ \text{let } x=au \\ x\rightarrow \infty \text{ this means } u\rightarrow \infty \\
\lim_{x\rightarrow \infty}\Big(1+\frac a{x}\Big)^x = \lim_{u\rightarrow \infty}\Big(1+\frac a{au}\Big)^{au}=\\ \lim_{u\rightarrow \infty} \Bigg( \Big(1+\frac 1{u}\Big)^u \Bigg)^a= \Bigg( \lim_{u\rightarrow \infty} \Big(1+\frac 1{u}\Big)^u \Bigg)^a=e^a
$$
Case 2 ($-\infty$, $-a/x$):
$$
\lim_{x\rightarrow -\infty}\Big(1-\frac a{x}\Big)^x \\ \text{let } x=-au \\ x\rightarrow -\infty \text{ this means } u\rightarrow \infty \\
\lim_{x\rightarrow -\infty}\Big(1-\frac a{x}\Big)^x = \lim_{u\rightarrow \infty}\Big(1+\frac {(-a)}{(-a)u}\Big)^{-au}=\\ \lim_{u\rightarrow \infty} \Bigg( \Big(1+\frac 1{u}\Big)^u \Bigg)^{-a}= \Bigg( \lim_{u\rightarrow \infty} \Big(1+\frac 1{u}\Big)^u \Bigg)^{-a}=e^{-a}
$$
But in case 3 & case 4, I do some thing that make things work but I feel it's wrong
Case 3 ($+\infty$, $-a/x$):
$$
\lim_{x\rightarrow \infty}\Big(1-\frac ax\Big)^x \\
\text{ let } x=-au \\
x\rightarrow \infty \text{ this means } u\rightarrow -\infty \\
\lim_{x\rightarrow \infty}\Big(1-\frac ax\Big)^x=\lim_{u\rightarrow -\infty}\Big(1+\frac {(-a)}{(-a)u}\Big)^{-au} = \lim_{u\rightarrow -\infty}\Big(1+\frac 1u\Big)^{-au} \\
\text{ let } y=\frac 1u \\
u\rightarrow -\infty \text{ this means } y\rightarrow 0\\
\text{ but actually } y\rightarrow 0^{-} \text { because } u\rightarrow -\infty \\
\text{ but let's take the general case that } y\rightarrow 0 \\
\lim_{u\rightarrow -\infty}\Big(1+\frac 1u\Big)^{-au} =\lim_{y\rightarrow 0}(1+y)^{-a(\frac 1y)} \\
\text{ now let } v=\frac 1y \\
y\rightarrow 0 \text{ this means } v\rightarrow \infty \\
\text{ but actually } v\rightarrow -\infty \text{ because } y\rightarrow 0^{-} \\
\text{ but let's take the case } v\rightarrow \infty \\
\text{ and this is where I feel what I'm doing is wrong} \\
\text{ I feel it's wrong to take the case } v\rightarrow \infty \\
\lim_{y\rightarrow 0}(1+y)^{-a(\frac 1y)} = \lim_{v\rightarrow \infty}\Big(1+\frac 1v \Big)^{-av} = \\ \lim_{v\rightarrow \infty}\Bigg[ \Big(1+\frac 1v \Big)^v \Bigg]^{-a} = \Bigg[ \lim_{v\rightarrow \infty} \Big(1+\frac 1v \Big)^v \Bigg]^{-a}=e^{-a}
$$
Case 4 ($-\infty$, $+a/x$):
$$
\lim_{x\rightarrow -\infty}\Big(1+\frac ax\Big)^x \\
\text{ let } x=au \\
x\rightarrow -\infty \text{ this means } u\rightarrow -\infty \\
\lim_{x\rightarrow -\infty}\Big(1+\frac ax\Big)^x=\lim_{u\rightarrow -\infty}\Big(1+\frac {a}{au}\Big)^{au} = \lim_{u\rightarrow -\infty}\Big(1+\frac 1u\Big)^{au} \\
\text{ let } y=\frac 1u \\
u\rightarrow -\infty \text{ this means } y\rightarrow 0\\
\text{ but actually } y\rightarrow 0^{-} \text { because } u\rightarrow -\infty \\
\text{ but let's take the general case that } y\rightarrow 0 \\
\lim_{u\rightarrow -\infty}\Big(1+\frac 1u\Big)^{au} =\lim_{y\rightarrow 0}(1+y)^{a(\frac 1y)} \\
\text{ now let } v=\frac 1y \\
y\rightarrow 0 \text{ this means } v\rightarrow \infty \\
\text{ but actually } v\rightarrow -\infty \text{ because } y\rightarrow 0^{-} \\
\text{ but let's take the case } v\rightarrow \infty \\
\text{ and this is where I feel what I'm doing is wrong} \\
\text{ I feel it's wrong to take the case } v\rightarrow \infty \\
\lim_{y\rightarrow 0}(1+y)^{a(\frac 1y)} = \lim_{v\rightarrow \infty}\Big(1+\frac 1v \Big)^{av} = \\ \lim_{v\rightarrow \infty}\Bigg[ \Big(1+\frac 1v \Big)^v \Bigg]^{a} = \Bigg[ \lim_{v\rightarrow \infty} \Big(1+\frac 1v \Big)^v \Bigg]^{a}=e^{a}
$$
I hope to find my answer with you guys
 A: Why not to use
$$y=\left(1+\frac{a}{x}\right)^x \implies \log(y)=x \log\left(1+\frac{a}{x}\right)$$ Now, using Taylor expansion
$$\log(y)=x \left(\frac{a}{x}-\frac{a^2}{2 x^2}+O\left(\frac{1}{x^3}\right)\right)=a-\frac{a^2}{2 x}+O\left(\frac{1}{x^2}\right)$$
$$y=e^{\log(y)}=e^a\left(1-\frac{a^2}{2 x} \right)+O\left(\frac{1}{x^2}\right)$$ The limit is the same when $x \to \pm \infty$; the only thing which would differ is the manner the limit is approached.
A: You don't really have four cases, rather two of them. Because
$$\lim_{x\to-\infty}\left(1\pm\frac ax\right)^x=\lim_{x\to\infty}\left(1\mp\frac{a}x\right)^{-x}=\lim_{x\to\infty}\left(1\mp\dfrac{-a}x\right)^{x}=\lim_{x\to\infty}\left(1\pm\dfrac{a}x\right)^{x}$$ and you needn't consider the limits to minus infinity.
Then as you did, with $a>0$,
$$\lim_{x\to\infty}\left(1+\frac ax\right)^x=\lim_{z\to\infty}\left(1+\frac 1z\right)^{az}=e^a$$
and using the substitution $x\to z+a$,
$$\lim_{x\to\infty}\left(1-\frac ax\right)^x=\lim_{z\to\infty}\left(1+\frac az\right)^{-z-a}=\lim_{z\to\infty}\left(1+\frac az\right)^{-z}\left(1+\frac az\right)^{-a}=e^{-a}$$
because the second factors tends to $1$.
