How to calculate limits of sequences?
Calculate the limits of the following sequences$$ a_n = \frac{n+4n-n^5}{n^3+3n^5-2n}$$$$ b_n = \sqrt{n^2+4} - \sqrt{n^2+2} $$
I know the formulas to transform sequences to find out its limit but I do not understand to which point one has to transform a sequence to see its limit. I mean the limit of $a_n$ seems to be $- \frac{1}{3}$ and the limit of $b_n$ seems to be $0$ by just looking at them. I did not calculate anything. So I wonder to which point I have to transform sequences to "see" their limits. Can anyone explain me what the goal is when it comes to calculate limits of sequences?
$\endgroup$ 43 Answers
$\begingroup$For the first one, divide the numerator and the denominator by the dominant power of the numerator that is $n^5$ in this case:$$\begin{align} \lim a_n &= \lim\frac{n+4n-n^5}{n^3+3n^5-2n} \\ &= \lim \frac{\color{blue}{5/n^3} - 1}{\color{blue}{1/n^2} + 3 - \color{blue}{2/n}} \\ &=\frac{-1}{3}\end{align}$$because all blue terms tend to $0$ as $n$ goes to infinity.
For the second one, here is the hint:$$b_n = \sqrt{n^2+4} - \sqrt{n^2+2} = \frac{2}{\sqrt{n^2+4} +\sqrt{n^2+2}}$$
$\endgroup$ $\begingroup$Numerically, the most dominant term in the numerator of $a_n$ is $-x^5$ and in the denominator it is $3x^5$, so the $$\lim_{n \to \infty} a_n=-\frac{1}{3}.$$
Next $$L=\lim_{n \to \infty}b_n= \lim_{n \to\infty}[\sqrt{n^2+4}-\sqrt{n^2+2}]= \lim_{n\to \infty} n[(1+4/n^2)^{1/2}-(1+2/n^2)^{1/2}].$$Use binomial expansion $(1+z)^k =1+kz+k(k-1)z^2/2+.... |z|<<1$Then $$L=\lim_{n\to \infty} n[1+2/n^2+...-1-1/n^2-...]=\lim_{n \to \infty}n/n^2= \lim_{n \to \infty}1/n=0$$
$\endgroup$ $\begingroup$For $b_n$, Consider $f(x)=\sqrt{x}$ on $[n^2+2, n^2+4]$. Note that $f$ satisfies the hypothesis of mean value theorem. Then there exist $c \in (n^2+2, n^2+4) $ such that $$\frac{1}{2\sqrt{c}}=f'(c)=\frac{\sqrt{n^2+4}-\sqrt{n^2+2}}{2}$$ So $$0<\sqrt{n^2+4}-\sqrt{n^2+2}=\frac{1}{\sqrt{c}}<\frac{1}{\sqrt{n^2+2}}$$The result follows by Squeeze theorem!
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