# Is $\lim_{n \to \infty} a_n ^{b_n} = e^{\lim_{n \to \infty}(a_n - 1)b_n}$ always true?

Consider $a_n$ and $b_n$ are two sequences which $\lim _{n \to \infty} a_n = 1$ and $\lim _{n \to \infty} b_n = \infty$ . Can we always use this formula ?

$$\lim_{n \to \infty} a_n ^{b_n} = e^{\lim_{n \to \infty}(a_n - 1)b_n}$$

Also, when can we use this method for functions ?

A famous case is $a_n = 1+ \frac{1}{n}$ and $b_n = n$ . So $\lim_{n \to \infty}(a_n - 1)b_n = 1$ and $a_n ^{b^n} = e^1 = e$

• The notation (of the right-hand side) is confusing. If you use two limits, make sure you let them go over different letters; as for now, what $n$ belongs to what limit? May 5 '17 at 6:11
• @vrugtehagel Yes , Thank you . I edited it . May 5 '17 at 6:13
• @JaideepKhare Your mean is L'Hôpital's rule ? May 5 '17 at 6:18
• @JaideepKhare No , I don't . May 5 '17 at 6:20
• @JaideepKhare Okay , thank you . May 5 '17 at 6:29

Yes, this formula can always be used.

Let's take a look at it derivation.It will be clear from the derivation that where it can be used.

$$\text{let}~~L= \lim_{n \to \infty} a_n ^{b_n}$$

$$\ln L= \lim_{n \to \infty} b_n \ln a_n$$ $$\ln L= \lim_{n \to \infty} b_n \ln (1+(a_n-1))$$

Since $a_n \to 1 \implies a_n-1 \to 0$ therefore, we can use the fact that : $$\lim_{x \to 0}\dfrac{\ln(1 + x)}{x}=1$$

We get

$$\ln L= \lim_{n \to \infty} b_n \left(\frac{\ln (1+a_n-1)}{a_n-1}\right)(a_n-1)=\underbrace {\lim_{n \to \infty} \left(\frac{\ln (1+a_n-1)}{a_n-1}\right)}_{=1} \cdot \lim_{n \to \infty} b_n(a_n-1)$$ $$\implies \ln L= \lim_{n \to \infty} b_n (a_n-1)$$ Hence $$L=e ^{\lim_{n \to \infty} b_n (a_n-1)}$$

• You have wrote $\ln L= \ln \lim_{n \to \infty} a_n ^{b_n}$ . Then $\ln L = \lim_{n \to \infty} \ln a_n ^{b_n}$ . Can you explain why it is true ? May 5 '17 at 6:33
• @Jaideep You have assumed the limit exists. Could you justify why/if this works if $\lim_{n\to\infty}a_n^{b_n} = \infty$? May 5 '17 at 6:36
• @adfriedman If the limit doesn't exist, then the limit in the exponent of $e$ will also not exist May 5 '17 at 6:50
• @S.H.W any finite number of operations or functions can be applied in any limit (which are well defined).Both in LHS and RHS. May 5 '17 at 6:54
• This is fine if $L\ne 0.$ But if $L=0$ then $\ln L$ does not exist, and $b_n(a_n-1) \to -\infty$ as $n\to \infty.$ E.g. if $a_n=1-1/n$ and $b_n=n^2.$ May 5 '17 at 8:49