# 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

## 5 Answers

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.

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 $$-\dfrac{1}{\sqrt{5}}$$.

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 $$\frac{-1}{\sqrt{5}} = \frac{-\sqrt{5}}{5}$$.

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, $$\lim_{x\rightarrow0}\frac{e^{3x}-e^{4x}}{\sqrt5\,x}=\frac1{\sqrt5}\lim_{x\rightarrow0}\left(3\frac{e^{3x}-1}{3x}-4\frac{e^{4x}-1}{4x}\right)=\frac1{\sqrt5}(3-4)=-\frac1{\sqrt5}$$

$$\lim_{x\to0}\frac{e^{3x}-e^{4x}}{\sqrt{5}x}$$

$$=\lim_{x\to0}\frac {e^{3x}(1-e^{x})}{\sqrt{5}x}$$

$$=\lim_{x\to0}\frac{-e^{3x}}{\sqrt5}\lim_{x\to0}\frac { (e^x-1)}{x}$$

$$=\dfrac{-1}{\sqrt5}$$