lim n→∞ n (e^a/n - e^b/n)
lim n→∞ n (e^a/n - e^b/n)
Answer
Explanation:
Step1: Let $x=\frac{1}{n}$, then as $n\rightarrow\infty$, $x\rightarrow0$.
The limit becomes $\lim_{x\rightarrow0}\frac{e^{ax}-e^{bx}}{x}$.
Step2: Apply L - H rule.
Since it is in the $\frac{0}{0}$ form ($e^{a\times0}-e^{b\times0}=1 - 1 = 0$), by L - H rule, if $\lim_{x\rightarrow c}\frac{f(x)}{g(x)}$ is in $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form, then $\lim_{x\rightarrow c}\frac{f(x)}{g(x)}=\lim_{x\rightarrow c}\frac{f'(x)}{g'(x)}$. Here $f(x)=e^{ax}-e^{bx}$, $f'(x)=ae^{ax}-be^{bx}$, $g(x)=x$, $g'(x) = 1$. So $\lim_{x\rightarrow0}\frac{ae^{ax}-be^{bx}}{1}$.
Step3: Evaluate the limit.
Substitute $x = 0$ into $\frac{ae^{ax}-be^{bx}}{1}$, we get $ae^{a\times0}-be^{b\times0}=a - b$.
Answer:
$a - b$