Question

One more limit question Im stuck on

Answer 1

\[\lim_{n \rightarrow \infty} \frac{ 10^{n+1} }{ 11^{n-1} }\]

Answer 2

Exponent rule, pulling out a 10 in the top,
and 1/11 in the bottom.\[\lim_{n\to\infty}\frac{10^{n+1}}{11^{n-1}}\quad=\quad \lim_{n\to\infty}\frac{10\cdot 10^n}{\frac{1}{11}\cdot11^{n}}\]
So umm something like this, ya?\[\large\rm =110\lim_{n\to\infty}\left(\frac{10}{11}\right)^n\]

Answer 3

If that first step was confusing,
then another way you can approach it is by multiplying top and bottom by 11,\[\large\rm \lim_{n\to\infty}\frac{10^{n+1}}{11^{n-1}}\quad=\quad \lim_{n\to\infty}\frac{10\cdot 10^n}{11^{n-1}}\cdot\frac{11}{11}\]The 11 in the bottom will get rid of the -1 in the exponent, ya?

Answer 4

Gotcha. Sorry for the late response. Got carried away studying and forgot to check back. That makes sense though @zepdrix, so this limit would go to zero?

Answer 5

Yes.