Is there other answer of $\lim\limits _{n\rightarrow \infty }\left( ( an+b)^{n}/( an+c)^{n}\right)$?

Question

Here is my thinking:
$ $$\displaystyle \begin{array}{{>{\displaystyle}l}} \ \ \ \ \ \ \lim\limits _{n\rightarrow \infty }\frac{( an+b)^{n}}{( an+c)^{n}} \ \ \ a\neq 0\\ =\ \lim\limits _{n\rightarrow \infty }\frac{e^{n\ln( an+b)}}{e^{n\ln( an+c)}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ =\ \lim\limits _{n\rightarrow \infty } e^{n\ln( an+b) -n\ln( an+c)} \ \\ =\ \lim\limits _{n\rightarrow \infty } e^{n[ \ln( an+b) -\ln( an+c)]}\\ =\lim\limits _{n\rightarrow \infty } e^{n\ln\left(\frac{an+b}{an+c}\right)} \ \ \ \ \ \ \ \ \ \ \ \ \\ =\ \lim\limits _{n\rightarrow \infty } e^{n\ln\left(\frac{1+\frac{b}{an}}{1+\frac{c}{an}}\right)} \ \ \ \ \ \ \ \ \\ =\lim\limits _{n\rightarrow \infty } e^{n\ln1} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ = 1 \end{array}$

But there are two standard solutions of the limit, one is based on $b=c$, the other is based on $b≠c$.
I don't know why to consider these situations. Whatever relationship between $b$ and $c$, seems to fit in my solution?
Please let me know where is wrong. Thanks for help

Answer 1

In your derivation this final step

$$\ldots=\ \lim\limits _{n\rightarrow \infty } e^{n\ln\left(\frac{1+\frac{b}{an}}{1+\frac{c}{an}}\right)} =\lim\limits _{n\rightarrow \infty } e^{n\ln1} \ldots$$

is uncorrect because you are taking the limit only for a single factor for the expression which is an indeterminate form.

To proceed, for $b\neq c$, as an alternative we have

$$\frac{( an+b)^{n}}{( an+c)^{n}}=\left(\frac{ an+b}{ an+c}\right)^n=\left(\frac{ an+c+b-c}{ an+c}\right)^n=$$

$$=\left(1+\frac{ b-c}{ an+c}\right)^n=\left[\left(1+\frac{ b-c}{ an+c}\right)^{an+c}\right]^{\frac n {an+c}}\to \left(e^{b-c}\right)^\frac1a=e^{\tfrac{b-c}{a}}$$

For the case $b=c$ we simply have

$$\frac{( an+b)^{n}}{( an+c)^{n}}=\frac{( an+b)^{n}}{( an+b)^{n}}=1$$

Refer also to the related

• Analyzing limits problem Calculus (tell me where I'm wrong).

Answer 2

Recall that $\lim_{n\rightarrow\infty}\big(1+\tfrac{x}{n}\big)^n=e^x$ for any $x\in\mathbb{R}$ (in fact this also works if $x$ is replaced by a complex sequence $z_n$ that converges to $z\in\mathbb{C}$). In any event,

$$\frac{(na+b)^n}{(na+c)^n}=\frac{(1+\tfrac{b/a}{n})^n}{(1+\tfrac{c/a}{n})^n}\xrightarrow{n\rightarrow\infty}e^{b/a}e^{-c/a}=e^{\tfrac{b-c}{a}}$$

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Commment: The classical limit $\lim_{n\rightarrow\infty}\big(1+\tfrac{x}{n}\big)^n=e^x$ can be established without resorting to L'Hospital theorem and without the introduction of logarithms. There are several discussions about this in MSE, here is example

Answer 3

Hint: To find $\ \lim\limits _{n\rightarrow \infty } e^{n\ln\left(\frac{1+\frac{b}{an}}{1+\frac{c}{an}}\right)} \ \ \ \ \ \ \ \ \\$ use the fact that $\ln (1+x) \sim x$ as $x \to 0$.