Evaluate $ab$ given the value of a limit If $$\lim_{x\to 0} \frac{ae^x -b}{x} = 2$$ then find $ab$.
I have tried everything that atleast I know about limits or algebra but I got not a single idea how to do this.
Please help
 A: So $\lim_{x\rightarrow 0}ae^{x}-b=0$, so $a=b$, then the expression is $2=a(e^{x})'|_{x=0}=a\cdot e^{0}=a$, so $ab=4$.
Note that having found that $a=b$, then $\lim_{x\rightarrow 0}\dfrac{ax-a}{x}=2$, so $a\lim_{x\rightarrow 0}\dfrac{e^{x}-1}{x}=2$. But $e^{0}=1$, so $a\lim_{x\rightarrow 0}\dfrac{e^{x}-e^{0}}{x-0}=2$. So the limit is exactly the form of the definition of derivative of $e^{x}$ at $x=0$, hence $a(e^{x})'|_{x=0}=2$. Note that the derivative of $e^{x}$ is itself, so $(e^{x})'|_{x=0}=e^{x}|_{x=0}=e^{0}=1$.
A: We have
$$e^x = \sum_{k = 0}^\infty\frac{x^k}{k!} = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$$
by Taylor Series expansion of $e^x$, so
$$\lim_{x\to 0} \frac{ae^x -b}{x} = \lim_{x\to 0}\frac{a(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...)-b}{x} = \lim_{x\to 0}\bigg(\frac{a}{x}+a+\frac{x}{2!}+...-\frac{b}{x}\bigg) = 2$$
Now, notice that as $x \to 0$, the terms $\frac{x}{2!}$, $\frac{x^2}{3!},...$ will all become $0$ and we are left with:
$$\lim_{x\to 0}\bigg(\frac{a-b}{x}+a\bigg) = 2$$
This is possible only when we have $a = b$ so that the limit $\lim_{x\to 0}\frac{a-b}{x}$ does not $\to \infty$ (notice that we have a real expression on RHS) and equal to $0$; and $a = 2$ so that the equation is satisfied. Therefore, $a = b = 2$ and $ab = 4$.
