Computing $\lim_{x\to 0}\frac{1}{x}(1-e^{\alpha x})$ using l'Hospital's rule

Question

I try to find the following limit:

$$ \underset{x\rightarrow0}{\lim}\frac{1}{x}\left(1-e^{\alpha x}\right).$$

I tried it by l'Hopital's rule but I am not sure of the calculations that I have made. Any solution or hints?

Answer 1

Consider $f(x)=e^{ax}$. The limit as x goes to 0 of $\frac{f(x)-0}{x-0}=\frac{e^{ax}-1}{x}$ is the derivtive of f at 0! And hence the limit you're looking for is $-f'(0)$.

Answer 2

Well the limit is indeterminate i.e., $\frac{0}{0}$ so we differentiate the top and bottom to obtain $$\frac{-\alpha e^{\alpha x}}{1}\to-\alpha,$$as $x\to 0$.

Answer 3

L'Hospital way is: $$\lim_{x\rightarrow0}\frac{1-e^{\alpha x}}x\stackrel{L'Hos.}=\lim_{x\rightarrow0}\frac{0-\alpha e^{\alpha x}}1=-\alpha$$ Another way to do is using taylor series: $$\lim_{x\rightarrow0}\frac{1-e^{\alpha x}}x=\lim_{x\rightarrow0}\frac{1-(1+(\alpha x)+(\alpha x)^2/2+...)}x=\lim_{x\rightarrow0}\frac{-(\alpha x)-(\alpha x)^2/2-...)}x=-\alpha$$

Answer 4

Let $f(x)=1-e^{\alpha x}$ and let $g(x)=x$. Then you are trying to figure out what $\lim_{x\to 0}h(x)$ is where $h(x)=f(x)/g(x)$. Clearly we may use l'Hospital's rule because $h(0)=f(0)/g(0)=0/0$, which means you are dealing with the indeterminate form $\frac{0}{0}$. Now consider that $f'(x)=-\alpha e^{\alpha x}$ and $g'(x)=1$. Thus, you have the following: $$ \underset{x\rightarrow0}{\lim}\frac{1}{x}\left(1-e^{\alpha x}\right)=\lim_{x\to 0}\frac{1-e^{\alpha x}}{x}\stackrel{\text{(LH)}}{=}\lim_{x\to 0}\frac{-\alpha e^{\alpha x}}{1}=\frac{-\alpha\cdot 1}{1}=-\alpha. $$