I'm trying to compute the following limit:
$$L=\lim_{x\to 0} \frac{e^{ax}-e^{bx}}{x} \tag{1}$$
And I have to use some of the following limits for it:
$$\lim_{x\to 0}(1+x)^{\frac{1}{x}}=e=\lim_{x\to \infty}\left(1+\frac{1}{x}\right)^{x}$$
I tried some substitutions, specially the first limit but I got only to:
$$L=\lim_{x\to 0} \frac{(x+1)^a-(x+1)^b}{x} \tag{2}$$
Which I tried to substitute $x$ for $x-1$ but this yielded nothing useful.
Can you give me a hint?
$$ L = \lim _{x\to 0}\left(\frac{e^{ax}-e^{bx}}{x}\right) $$ $$ L = \lim _{x\to 0}\left(\frac{e^{ax}\left(1 - e^{(b-a)x}\right)}{x}\right) $$ $$ L = \lim _{x\to 0}\left(e^{ax} \cdot \frac{1 - e^{(b-a)x}}{x}\right) $$
Notice that as $x \to 0, e^{ax}\to 1$, so we can focus on the second part of the product:
$$ L = \lim _{x\to 0}\left(\frac{1 - e^{(b-a)x}}{x}\right) $$
Now, let's rewrite $1 - e^{(b-a)x}$ to look more like the provided limits. Use the substitution $$h = (b-a)x$$ As $x \to 0, h \to 0$
$$ L = \lim _{h\to 0}\left(\frac{1 - e^h}{\frac{h}{b-a}}\right)$$ $$ L = (b-a)\lim _{h\to 0}\left(\frac{1 - e^h}{h}\right) $$ $$ L = -(b-a)\lim _{h\to 0}\left(\frac{ e^h - 1}{h}\right) $$
Let $u=e^h - 1$. Then $e^h = 1+u$ implying $h=\ln(1+u)$. As $h \to 0, u \to 0$ since $1+u \to 1$
$$ \lim _{h\to 0}\left(\frac{ e^h - 1}{h}\right) = \lim _{u\to 0}\left(\frac{1+u- 1}{\ln(1+u)}\right)$$ $$ = \lim _{u\to 0}\left(\frac{u}{\ln(1+u)}\right)$$ $$ = \lim _{u\to 0}\left(\frac{1}{(\frac{1}{u})\ln(1+u)}\right)$$ $$ = \lim _{u\to 0}\left(\frac{1}{\ln((1+u)^\frac{1}{u})}\right)$$
Let $u=\frac{1}{n}$, so $$ \lim _{h\to 0}\left(\frac{ e^h - 1}{h}\right) = \lim _{u\to 0}\left(\frac{1}{\ln((1+u)^\frac{1}{u})}\right) $$ $$ = \lim _{n\to \infty}\left(\frac{1}{\ln((1 + \frac{1}{n})^n)}\right) $$ $$ = \frac{1}{\ln(e)} $$ $$ = 1 $$
So $$ L = -(b-a)\lim _{h\to 0}\left(\frac{ e^h - 1}{h}\right) $$ $$ = -(b-a) $$ $$ = a-b $$