Simplifying and taking the limit

Question

I would like to compute the following limit $$\lim_{n\to\infty}\frac{10^{n^{2}}}{10^{(n+1)^{2}}}$$ but I'm having a hard time simplifying. Can anyone explain to me the properties of these exponents?

Thanks

Answer 1

Notice the following:

$\frac{10^{n^{2}}}{10^{(n+1)^{2}}}$=$\frac{10^{n^{2}}}{10^{n^2+2n+1}}$

$\frac{10^{n^{2}}}{10^{n^2+2n+1}}=\frac{10^{n^{2}}}{10^{n^2}10^{2n}10^{1}}$

$\frac{10^{n^{2}}}{10^{n^2}10^{2n}10^{1}}$=$\frac{1}{10^{2n+1}}$

So, I hope now, it should be obvious what happens as $n$ gets infinitely large.

Answer 2

Hint: $$10^{(n+1)^2}=10^{n^2+2n+1}=10^{n^2}10^{2n+1}.$$

Answer 3

First, you can simplify this fraction: $$\frac{10^{n^2}}{10^{(n+1)^2}}=\frac{10^{n^2}}{10^{n^2+2n+1}}$$ Here, you can use the property of exponents which states: $$\frac{a^m}{a^n}=a^{m-n}$$ Applying the property, we get: $${10^{n^2-(n^2+2n+1)}}={10^{-(2n+1)}}=\frac1{10^{(2n+1)}}$$ When we input infinity, we get: $$\frac1{10^{2\infty+1}}=\frac1{10^\infty}=\frac1\infty=0$$

Answer 4

$$\lim_{n\to\infty}\frac{10^{n^{2}}}{10^{(n+1)^{2}}}=\lim_{n\to\infty}\frac{10^{n^{2}}}{10^{n^2}\cdot10^{2n+1}}=\lim_{n\to\infty}\frac{1}{10^{2n+1}}=0$$