MathB.in
New
Demo
Tutorial
About
Compute the following limit: \begin{align*} \lim_{t \to \infty}\bigg(\frac{3^{t}+5^{t}}{2}\bigg)^\frac{1}{t} \end{align*} Solution. We can see from some experience, that this limit will be an indeterminate form: $\infty^{0}$. Problematic. But, we also know that when the variable is in the power, taking the $ln$ of the limit will also help to deal with that. Let: \begin{align*} L & = \lim_{t \to \infty}\bigg(\frac{3^{t}+5^{t}}{2}\bigg)^\frac{1}{t} \end{align*} Take the $ln$ of both sides and continue: \begin{align*} ln(L) & = ln\bigg[\lim_{t \to \infty}\bigg(\frac{3^{t}+5^{t}}{2}\bigg)^\frac{1}{t}\bigg] \\\\ & = \lim_{t \to \infty}ln\bigg[\bigg(\frac{3^{t}+5^{t}}{2}\bigg)^\frac{1}{t}\bigg] & \text{(property of limits)}\\\\ & = \lim_{t \to \infty}\frac{1}{t}ln\bigg(\frac{3^{t}+5^{t}}{2}\bigg) & \text{(property of logarithms)}\\\\ & = \lim_{t \to \infty}\frac{1}{t}\bigg[ln(3^{t}+5^{t})-ln2\bigg] & \text{(property of logarithms)}\\\\ & = \lim_{t \to \infty}\frac{1}{t}ln(3^{t}+5^{t})-\lim_{t\to\infty}\frac{1}{t}ln(2) & \text{(distribute)}\\\\ & = \lim_{t \to \infty}\frac{1}{t}ln(3^{t}+5^{t}) & \text{(simplify, second limit goes to zero)}\\\\ & = \lim_{t \to \infty}\frac{ln(3^{t}+5^{t})}{t} & \text{(rewrite)}\\ \end{align*} Another tough limit, but we're getting there. We purposefully rewrite the limit so we can get another indeterminate form: $\frac{\infty}{\infty}$. L'Hopital's Rule is needed, as there is not really an algebraic way to deal with this: \begin{align*} ln(L) & = \lim_{t\to\infty}\frac{ln(3^{t}+5^{t})}{t}\\\\ & = \lim_{t\to\infty}\frac{\frac{d}{dt}ln(3^{t}+5^{t})}{\frac{d}{dt}t} & \text{(Apply L'Hopital's Rule)}\\\\ & = \lim_{t\to\infty}\ \frac{1}{3^{t}+5^{t}}\cdot(3^{t}ln3 + 5^{t}ln5) & \text{(differentiate)}\\\\ & = \lim_{t\to\infty}\frac{3^{t}ln3 + 5^{t}ln5}{3^{t}+5^{t}} & \text{(rewrite)}\\ \end{align*} At this point, it's still not looking much better, but we can apply some intuition. With credit to blackpenredpen, wouldn't it be nice if we could do some sort of leading coefficient test on that fraction? Intuition (and a graph) tells us that the $5^{t}$ dominates the numerator and denominator. If this is true, doing some sort of leading coefficient test on $\frac{5^{t}ln5 + 3^{t}ln3}{5^{t}+3^{t}}$ would reduce to $\frac{5^{t}ln5}{5^{t}}$, which would leave us with $ln(5)$. More algebra is needed to show this. Let's start with a factoring trick to eliminate the $3^{t}$ terms: \begin{align*} ln(L) & = \lim_{t\to\infty}\frac{3^{t}ln3 + 5^{t}ln5}{3^{t}+5^{t}} \\\\ & = \lim_{t\to\infty}\frac{3^{t}(ln3+\frac{5^{t}}{3^{t}}ln5)}{3^{t}(1+\frac{5^{t}}{3^{t}})} & \text{(factor)} \\\\ & = \lim_{t\to\infty}\frac{(ln3+\frac{5^{t}}{3^{t}}ln5)}{(1+\frac{5^{t}}{3^{t}})} & \text{(cancel $3^{t}$ terms)} \\\\ & = \lim_{t\to\infty}\frac{ln3+ln5(\frac{5}{3})^{t}}{1+(\frac{5}{3})^{t}} & \text{(rearrange, rewrite)} \\ \end{align*} So close. Those constant terms in the numerator and denominator won't mean much as $t\to\infty$, so our intuition about $ln5$ being the limit looks to be solid. If we try to work this limit, we get another indeterminate $\frac{\infty}{\infty}$, so we can apply L'Hopital's Rule again. Note what happens to those constant terms - they are negligible, and differentiating actually deals with that! \begin{align*} ln(L) & = \lim_{t\to\infty}\frac{ln3+ln5(\frac{5}{3})^{t}}{1+(\frac{5}{3})^{t}}\\\\ & = \lim_{t\to\infty}\frac{\frac{d}{dt}\bigg[ln3+ln5(\frac{5}{3})^{t}\bigg]}{\frac{d}{dt}\bigg[1+(\frac{5}{3})^{t}\bigg]} & \text{(Apply L'Hopital's Rule)} \\\\ & = \lim_{t\to\infty}\frac{(\frac{5}{3})^{t}\cdot ln(\frac{5}{3})\cdot ln5}{(\frac{5}{3})^{t}\cdot ln(\frac{5}{3})} & \text{(differentiate)} \\\\ & = \lim_{t\to\infty} ln5 & \text{(simplify!)} \\\\ ln(L) &= ln5 & \text{(take the limit of a constant)} \\\\ L & = 5 & \text{(simplify)} \end{align*} So: \begin{align*} \lim_{t \to \infty}\bigg(\frac{3^{t}+5^{t}}{2}\bigg)^\frac{1}{t} = 5\\ \end{align*} Done.
ERROR: JavaScript must be enabled to render input!
Wed, 20 Feb 2019 21:18 GMT