Why $\lim \limits_ {n\to \infty}\left (\frac{n+3}{n+4}\right)^n \neq 1$?

Solution 1:

Because $1^\infty$ is a tricky beast. Perhaps the power overwhelms the quantity that's just bigger than $1$, but approaching $1$, and the entire expression is large. Or perhaps not...

Perhaps the power overwhelms the quantity that's just smaller than $1$, but approaching $1$, and the entire expression tends to $0$ . Or perhaps not...

In your case, $$ {n+3\over n+4} = 1-{1\over n+4}. $$ And, as one can show (as you did): $$\lim\limits_{n\rightarrow\infty}(1-\textstyle{1\over n+4})^n = \lim\limits_{n\rightarrow\infty}\Bigl[ (1-\textstyle{1\over n+4})^{n+4}\cdot (1-{1\over n+4})^{-4}\Bigr] = e^{-1}\cdot1=e^{-1}.$$

Here, the convergence of $1-{1\over n+4}$ to 1 is too fast for the $n^{\rm th}$ power to drive it back down to $0$.

Solution 2:

Another way to see:

$\left(\frac{n+3}{n+4}\right)^n=\left(\frac{1+\frac{3}{n}}{1+\frac{4}{n}}\right)^n=\frac{\left(1+\frac{3}{n}\right)^n}{\left(1+\frac{4}{n}\right)^n}.$

Since, $\lim_{n\to\infty}\left(1+\frac{c}{n}\right)^n=e^c$, follows

$$\lim_{n\to\infty}\left(\frac{n+3}{n+4}\right)^n=\lim_{n\to\infty}\frac{\left(1+\frac{3}{n}\right)^n}{\left(1+\frac{4}{n}\right)^n}=\frac{e^3}{e^4}=e^{-1}. $$

Solution 3:

As David Mitra makes clear in his answer, there is a more fundamental question underlying your question, which is:

• why is $\displaystyle \lim_{n\to \infty} (1 + 1/n)^n$ not equal to $1$?

The same (fallacious) reasoning as the one you give applies. However, this limit is known to equal $e$, not $1$ (and your limit of $e^{-1}$ can easily be obtained from that limit, as David essentially shows).

As David says, there is a tension between the term in parentheses which is tending to $1$ from above, and the power of $n$, which, when applied to any fixed number $>1$, will give larger and larger answers as $n$ increases.

You might want to consider some other related limits to see how they behave, e.g.:

• $\displaystyle\lim_{n\to\infty} (1+1/n^2)^n$ .

• $\displaystyle \lim_{n\to\infty} (1 + 1/\log n)^n$ .

Solution 4:

While intuitively it seems "obvious" that $1^\infty$ has to be one, let me point to you that $ \dfrac{n+3}{n+4} <1$. Then the meaning of

$$\left(\frac{n+3}{n+4}\right)^n $$

is a number strictly less than $1$ to a huge power. Now, a number between $0$ and $1$ raised to a larger power gets smaller and smaller, so is it still obvious that it approaches 1?