i'm trying to find the following limit, if it exists, $$\lim_{n→ ∞} \frac{(n+7)^{n-5}}{n^n}$$.

Now, I've tried division like $$\lim_{n→ ∞}\frac{|n+1|}{|n|}$$, or dividing by the highest power, but I can't solve it.

Also thought of converting it to a function and using L'Hôpital's rule, but that seems more complicated than it should be. I think it's a simple limit to solve, but somehow I'm stuck and can't think of a way to tackle it.

Can someone give a hint or method; even a complete solution, but explaining the thought process?


Usually the method is to look for similariarity to patterns you've seen before and know the limits of. In this case, there is a limit $n\to \infty$ , there are exponents $n$ and a fraction and this immediately reminds me of the definition of Euler's constant $e$ and generally the term $e^x$.

The definition of $e$ I have in mind is

$$e = \lim_{n\to \infty}\left(1+\frac1n\right)^n$$

In order to get the limit at hand in that direction, we manipulate it to have a fraction term where both enumerator and denominator have exponent $n$, then take the exponent out of the fraction:

$$\frac{(n+7)^{n-5}}{n^n} = \frac{(n+7)^{n}}{n^n}\frac1{(n+7)^5} = \left(\frac{n+7}n\right)^n\frac1{(n+7)^5} = \left(1+\frac7n\right)^n\frac1{(n+7)^5}$$

This is just a manipulation of terms, but with a goal. They key insight is that if we now go to the limit

$$\lim_{n\to\infty}\frac{(n+7)^{n-5}}{n^n} = \lim_{n\to\infty}\left[\left(1+\frac7n\right)^n\frac1{(n+7)^5}\right],$$

the first term on the right hand side goes to $e^7$:

$$\lim_{n\to\infty}\left(1+\frac7n\right)^n = e^7$$

You can find this limit for example here: https://en.wikipedia.org/wiki/Exponential_function

The second part of the right hand side is much easier and elementary: $$\lim_{n\to\infty}\frac1{(n+7)^5} = 0.$$

So we finally can apply the theorem that the limit of a product is the product of the limits (if they exist and are not $\infty$):

$$\lim_{n\to\infty}\frac{(n+7)^{n-5}}{n^n} = \lim_{n\to\infty}\left[\left(1+\frac7n\right)^n\frac1{(n+7)^5}\right] = \lim_{n\to\infty}\left(1+\frac7n\right)^n \lim_{n\to\infty}\frac1{(n+7)^5} = e^7\times0 = 0$$


Write it as

$$\left(\frac{n+7}{n}\right)^{n-5} \frac{1}{n^5}.$$

The first factor has a finite limit, which you can get by the log trick. The second factor goes to zero. So the answer is $0.$

  • $\begingroup$ Ahh, didn't even cross my mind! Thanks both for the answer and the explanation $\endgroup$ – sdds Feb 18 '19 at 19:21

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