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I tried this way : $$\lim_{n\to \infty}\left(\frac{a+n}{b+n}\right)^n=\lim_{n\to \infty}\left(1+\frac{a-b}{b+n}\right)^n$$

We know $\lim_{n\to \infty}\left(1+\frac1u\right)^u=e$. therefor in this case we have: $$\lim_{n\to \infty}\left(1+\frac{a-b}{b+n}\right)^{\tfrac{b+n}{a-b}}=e$$ Here $n$ goes to infinity so we can ignore the number ($b$)added to it in numerator of the exponent. So $\lim_{n\to \infty}\left(1+\frac{a-b}{b+n}\right)^n=e^{a-b}$

Is my answer right? and is there any other approach to evaluate the limit?

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  • $\begingroup$ You cannot have $a-b$ in the denominator if $a=b$ so that's a separate case but $e^{a-b}=1$ is also correct in that case. $\endgroup$
    – Neat Math
    Commented Oct 31, 2020 at 9:56

2 Answers 2

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I think it is easier to write the expression as $$\frac{(1+\frac{a}{n})^{n}}{(1+\frac{b}{n})^{n}}$$ and take it from there.

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  • $\begingroup$ Yes. I wish I could accept two answers ;) $\endgroup$
    – Etemon
    Commented Oct 31, 2020 at 10:23
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    $\begingroup$ @soheil The other answer takes you from where you'd got to to the solution, which is a good thing. But in fact I think it's flaky unless you have a lot of machinery at your disposal ( I may be wrong). $\endgroup$ Commented Oct 31, 2020 at 10:43
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Your answer is right but, following your idea, we should proceed as follows

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

For a rigorous justification of this step you can refer to the following

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    $\begingroup$ I am not sure just how straightforward it is going to be to evaluate the (simultaneous) limit of the r.h. expression. The exponential limit $(1+\frac{x}{n})^n\to e^x$ is "elementary", replacing integral $n$ by arbitrary real $\alpha$ less so; and then there's the outer exponent - I'd need to take logs to make sure that the two limits interact in the way they in fact do, again a bit deeper than the simple exponential limit. $\endgroup$ Commented Oct 31, 2020 at 10:41
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    $\begingroup$ I agree with what you say, but to prove it you can't take the limits separately without a bit of argument. $\endgroup$ Commented Oct 31, 2020 at 14:27
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    $\begingroup$ @ancientmathematician I've added some reference. The first two are useful in general also with others approch. $\endgroup$
    – user
    Commented Nov 1, 2020 at 7:28
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    $\begingroup$ Thanks. I'm glad I wasn't missing anything, and that to follow this line we need the continuity of $\exp$ and the existence of the $\log$ function. To be honest, without knowing at exactly what point of a real analysis course the OP has reached it is hard to answer a question like this. $\endgroup$ Commented Nov 1, 2020 at 9:14
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    $\begingroup$ @ancientmathematician Yes I agree, often this is an issue! Anyway give some good reference is always useful. Bye $\endgroup$
    – user
    Commented Nov 1, 2020 at 9:22

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