If $\lim_{x\to 0}\frac{ae^x-b}{x}=2$ the find $a,b$ 
If $$\lim_{x\to 0}\frac{ae^x-b}{x}=2$$ the find $a,b$

$$
\lim_{x\to 0}\frac{ae^x-b}{x}=\lim_{x\to 0}\frac{a(e^x-1)+a-b}{x}=\lim_{x\to 0}\frac{a(e^x-1)}{x}+\lim_{x\to 0}\frac{a-b}{x}=\boxed{a+\lim_{x\to 0}\frac{a-b}{x}=2}\\
\lim_{x\to 0}\frac{a-b}{x} \text{ must be finite}\implies \boxed{a=b}\\
$$
Now I think I am stuck, how do I proceed ?
 A: You are almost done. $a+0=2$.
A: Going off of your work, we know that $a=b$. Thus, the question is: what $a$ makes
$$\lim_{x\to 0}a\frac{e^x-1}{x}=2?$$
Using the Taylor Series for $e^x$, we see
$$2=\lim_{x\to 0}a\frac{e^x-1}{x}=\lim_{x\to 0}a\frac{\sum_{n=0}^\infty\frac{x^n}{n!}-1}{x}=\lim_{x\to 0}a\frac{\sum_{n=1}^\infty\frac{x^n}{n!}}{x}$$
$$=\lim_{x\to 0}a\sum_{n=1}^\infty\frac{x^{n-1}}{n!}=a+\lim_{x\to 0}x\left(\sum_{n=2}^\infty\frac{x^{n-2}}{n!}\right)=a.$$
Thus, $a=b=2$
A: If you plug $a=b$ into your starting expression you get
$$\lim_{x\to 0}a\frac{e^x-1}{x} = a \left.\left(e^x \right)'\right|_{x=0}= a \stackrel{!}{=} 2$$
Btw, you can show $a=b$ a bit easier by noting that
$$\lim_{x\to 0}\frac{ae^x-b}{x}= 2 \Rightarrow \lim_{x\to 0}\left( \frac{ae^x-b}{x}\cdot x\right) = \lim_{x\to 0}\left(ae^x-b\right) = 0$$
$$\Rightarrow a-b = 0$$
A: You can use the Taylor series, $e^x=1+x+$ terms of order $x^2$ and higher.  Plug that in.  The fact that $a=b$ cancels the $1$ and you will be working with the $x$ term.
A: Option:
1) $\lim_{x \rightarrow 0}\dfrac{ae^x-b}{x}=2$;
$\lim_{x \rightarrow 0}(ae^x-b)=$
$\lim_{x \rightarrow \infty}((\dfrac {ae^x-b}{x})\cdot x)=$
$\lim_{x \rightarrow 0}\dfrac{ae^x-b}{x} \cdot \lim_{x \rightarrow 0} x=$
$2 \cdot 0=0$; 
$\rightarrow a=b$;
2) $\lim_{x \rightarrow 0} a(\dfrac{e^x-1}{x})=2$;
$a=?$
