Vtgyjfy

I am working with a 12 year old learning about limits, and am having a tough time proving

lim$_{n rightarrow infty}$ (1 - $frac{3}{n-2})^n$ = $frac{1}{e^3}$

using

lim$_{n rightarrow infty}$ (1 + $frac{x}{n})^n$ = lim$_{n rightarrow 0}$ (1 + $n$)$^{frac{x}{n}}$ = $e^x$.

I am probably missing some small analysis trick.

one thing I tried is that the negative 3 is why its $e^{-3}$, but where does the $n-2$ come from, how can I account for that?

ALSO, need helps showing

lim$_{n rightarrow infty}$ ($frac{n+2}{n-3})^n$ = $e^5$

would you rewrite $frac{n+2}{n-3}$ = $frac{n}{n-3}$ + $frac{2}{n-3}$??

For the first one, take logarithms and note that
$$lim_{n to infty} n log(1 - frac{3}{n-2}) = lim_{n to infty} frac{n}{n-2} cdot lim_{n to infty} (n-2) log(1 - frac{3}{n-2}).$$
The second limit on the right-hand side can be computed using your provided fact.

For the other question, Arthur has already given you a hint in the comments.

You could rewrite the limit:

$$lim_{n to infty}left(1-frac{3}{n-2}right)^n = lim_{n to infty}left(1+frac{-3}{n-2}right)^n = lim_{n to infty}left[color{blue}{left(1+frac{-3}{n-2}right)^{n-2}}right]^{color{green}{frac{n}{n-2}}}$$

Now, just make use of the standard limit $color{blue}{e^x = lim_limits{n to infty}left(1+frac{x}{n}right)^n}$. Since $color{green}{frac{n}{n-2} to 1}$ as $n to infty$, the limit becomes $left(color{blue}{e^{-3}}right)^{color{green}{1}} = e^{-3} = frac{1}{e^3}$.

For your second question, I think it would be the simplest to write $frac{n+2}{n-3}$ as $1+frac{5}{n-3}$. The limit is therefore rewritten:

$$lim_{n to infty}left(frac{n+2}{n-3}right)^n = lim_{n to infty}left(1+frac{5}{n-3}right)^n = lim_{n to infty}left[color{blue}{left(1+frac{5}{n-3}right)^{n-3}}right]^{color{green}{frac{n}{n-3}}}$$

The exact same idea applies here.

Just take $m=n-2$ and $x=-3$. Then you have $lim_{mrightarrow infty}(1+frac x m)^{m+2}$. You can then write that as $lim_{mrightarrow infty}(1+frac x m)^{m}(1+frac x m)^2$. If you can show that you can split that into separate limits, then you have ($lim_{mrightarrow infty}(1+frac x m)^{m}$) ($lim_{mrightarrow infty}(1+frac x m)^{2}$). The right parentheses clearly goes to $1$, and the left parentheses clearly goes to $e^x$. Substitute $-3$ in for $x$, and QED.




For your second limit, take $m=n-3$. Then $frac {n+2}{n-3}=frac {m+5} m=1+frac5 m$. Proceed similarly.




Also, if it helps you get a more intuitive idea why $frac {m+5} {m}$ leads to a limit of $e^5$, consider that the statement $lim_{nrightarrow infty}(1+frac 1 n )^{n}=e$ can be written as $lim_{nrightarrow infty}(frac {n+1} n)^{n}=e$. Then $lim_{nrightarrow infty}(frac {n+2}{n} )^{n}=lim_{nrightarrow infty}left(frac {n+2}{n+1} cdot frac {n+1}{n}right)^{n}$. Both $frac{n+2}{n+1}$ and $frac {n+1}{n}$ lead to a limit of $e$, so together they yield a limit of $e^2$. Similiary, $frac{n+5}n$ yields a limit of $e^5$.