I am trying to find $\lim \limits_{n \to \infty} \frac{2^n + 3^n} {10^{n \over 2}}$ and am getting that it does not exist when it should be zero. Here is my work:
$$\lim \limits_{n \to \infty} \frac{2^n + 3^n} {10^{n \over 2}} = \lim \limits_{n \to \infty} \frac {2^n}{10^{n \over 2}}+ \lim \limits_{n \to \infty}\frac {3^n}{10^{n \over 2}}= \lim \limits_{n \to \infty} {(\frac {2} {10}})^n10^{n \over 2} + \lim \limits_{n \to \infty} {(\frac {3} {10}})^n10^{n \over 2}= \lim \limits_{n \to \infty} e^{\ln({(\frac {2} {10}})^n10^{n \over 2})}+\lim \limits_{n \to \infty}e^{\ln({(\frac {3} {10}})^n10^{n \over 2})}= \lim \limits_{n \to \infty} e^{\ln({(\frac {2} {10}})^n)+ \ln(10^{n \over 2})}+\lim \limits_{n \to \infty} e^{\ln({(\frac {3} {10}})^n)+ \ln(10^{n \over 2})}= \lim \limits_{n \to \infty} e^{n\ln({\frac {2} {10}})+ {n \over 2}\ln(10)}+\lim \limits_{n \to \infty} e^{n\ln({\frac {3} {10}})+ {n \over 2}\ln(10)}$$At this point, I am seeing that each of these limits approach infinity as the exponent of $e$ gets arbitrarily large, so the entire limit does not exist. Where did I go wrong?
Why is $\lim \limits_{n \to \infty} \frac{2^n + 3^n} {10^{n \over 2}}$ $0$ and not $\infty$
calculuslimits
Best Answer
Your work, while a bit unnecessarily long, actually looks completely fine. At the end, you get the limit as $$\lim \limits_{n \to \infty} e^{n\ln({\frac {2} {10}})+ {n \over 2}\ln(10)}+\lim \limits_{n \to \infty} e^{n\ln({\frac {3} {10}})+ {n \over 2}\ln(10)}$$
which is right. But $\ln(2/10)+\frac{1}{2}\ln(10)$ and $\ln(3/10)+\frac{1}{2}\ln(10)$ are both negative numbers. So the exponents both end up going to $-\infty$.
Also, an easier way might be to bound the limit function like $$0\le \frac{2^n+3^n}{10^{n/2}}\le \frac{3^n+3^n}{10^{n/2}}=2\cdot \left(\frac{3}{\sqrt{10}}\right)^n\to 0$$ since $\frac{3}{\sqrt{10}}<1$.