Find the limit of sequence without using derivatives

Question

$$\lim_{n\to\infty}\large n(a^\frac{1}{n}-1) 0$$ $$a>0$$

is the problem. So far what I tried was transforming $$(a^\frac{1}{n}-1)$$

by using the formula $x^n-1 = (x-1)(x^{n-1} + x^{n-2} +...+1)$ but I fail to see how this would help me in this instance, or I just may be doing it wrong because it's the nth root that is giving me problems. Any help would be appreciated

score 0 · Answer 1 · answered Dec 08 '20 at 10:48

0

$$L=\lim_{n\to \infty} n(a^1/n-1)$$ Let $t=1/n$, then $$L=\lim_{t\to 0} \frac{a^t-1}{t}=\lim_{t \to 0} \frac{1+t\ln a+(t\ln a)^2/2+...-1}{t}=\ln a, a>0$$

answered Dec 08 '20 at 10:48

Z Ahmed

43,235

Thanks for your comment. Can you please clarify how did you go from $a^t - 1$ to $(1+t\ln(a) + ....-1)$? – john doe Dec 08 '20 at 10:59
$a^t=e^{\ln a^t}= e^{t \ln a}$, then use $e^z=1+z+z^2/2!+z^3/3!+...$ – Z Ahmed Dec 08 '20 at 11:07

Find the limit of sequence without using derivatives

1 Answers1