the limit of the function $(e^{3x}-e^{4x})/(\sqrt{5}x)$ as $x\to0$ How to find the limit of the function
$$\lim_{x\to0}\frac{e^{3x}-e^{4x}}{\sqrt{5}x}$$
I know that L'hopital's rule can be used to get the answer of $\dfrac{-\sqrt{5}}{5}$, but I tried to do it without the L'hopital and i dont how to solve it
 A: Use $e^x=1+x+...$, where you neglect all the other terms because you are interested in the limit of small $x$. In your case: $e^{4x}=1+4x+...$ and $e^{3x}=1+3x+...$. Therefore, the numerator is $e^{3x}-e^{4x}=-x+...$. Just simplify with the x in the denominator and you get the result.
A: Using the well-known result:
$$\lim_{x\to 0} \frac{a^x-1}{x}=\ln a$$
we have:
$$\lim_{x\to0}\frac{e^{3x}-e^{4x}}{x}=\lim_{x\to0}\frac{e^{3x}-1}{x}-\lim_{x\to0}\frac{e^{4x}-1}{x}=3-4=-1$$
and your limit is thus $-\dfrac{1}{\sqrt{5}}$.
A: Consider rewriting the inside of the limit as 
$$\frac{\frac{e^{3x}}{x}-\frac{e^{4x}}{x}}{\sqrt{5}}$$
Then write $e^{3x}$ and $e^{4x}$ in Taylor series form which gives,
$$\frac{e^{3x}}{x} = \sum_{n=0}^{\infty}\frac{3^{n}x^{n-1}}{n!}$$
$$\frac{e^{4x}}{x} = \sum_{n=0}^{\infty}\frac{4^{n}x^{n-1}}{n!}$$
The troublesome terms are when $n = 0$ and we have $1/x$ which will blow up as we take the limit. But luckily, we have $1/x$ from the first series and we take away $1/x$ from the second series, so this troublesome term disappears.
What we are left with are the two constant terms when $n=1$ which result in $3-4 = -1$.
All other terms of the series involve an $x$ to a positive power and so will disappear in the limit.
So we finally are left with $\frac{-1}{\sqrt{5}} = \frac{-\sqrt{5}}{5}$.
A: $$\lim_{x\to0}\frac{e^{3x}-e^{4x}}{\sqrt{5}x}$$
$$=\lim_{x\to0}\frac {e^{3x}(1-e^{x})}{\sqrt{5}x}$$
$$=\lim_{x\to0}\frac{-e^{3x}}{\sqrt5}\lim_{x\to0}\frac { (e^x-1)}{x}$$
$$=\dfrac{-1}{\sqrt5}$$
A: Let's start out by defining exponential and logarithmic functions via
$$\ln y=\int_1^y\frac1tdt$$
For $t\in[1,y]$, $1/y\le1/t\le1$ so
$$\frac{y-1}y=\int_1^y\frac1ydt\le\int_1^y\frac1tdt=\ln y\le\int_1^y1dt=y-1$$
So we have
$$1\le\frac{y-1}{\ln y}\le y$$
By the squeeze theorem
$$\lim_{y\rightarrow1^+}\frac{y-1}{\ln y}=1$$
The same argument works for the left hand limit as well so let $x=\ln y$ and then
$$\lim_{x\rightarrow0}\frac{e^x-1}x=\lim_{y\rightarrow1}\frac{y-1}{\ln y}=1$$
With that limit established,
$$\lim_{x\rightarrow0}\frac{e^{3x}-e^{4x}}{\sqrt5\,x}=\frac1{\sqrt5}\lim_{x\rightarrow0}\left(3\frac{e^{3x}-1}{3x}-4\frac{e^{4x}-1}{4x}\right)=\frac1{\sqrt5}(3-4)=-\frac1{\sqrt5}$$
