Confusion in a limit problem?
- Thread starter meer
- Start date
I'm having a bit of trouble reading your handwriting, but here's what I think your working shows.
Assuming that the above is correct, then I agree with all of your workings, and the answer is correct. The small bit of working not shown is that the limit of a fraction is equal to the limit of its numerator divided by the limit of its denominator, excepting cases of indeterminate forms. That is to say:
\(\displaystyle \displaystyle \lim_{n \to \infty}{\frac{1+n^{-1/6}}{n^{1/2}+2n^{1/6}}}=\frac{\lim_{n \to \infty}({1+n^{-1/6}})}{\lim_{n \to \infty}{(n^{1/2}+2n^{1/6})}}=\frac{1}{\infty}=0\)
As to your confusion about whether to divide by sqrt(n) or n, the general wisdom is, as you noted, to divide by the highest power in the denominator. But, as you've also demonstrated, this is not a hard and fast rule. You don't always have to follow it exactly, although sometimes it means extra work if you don't. If you'd begun by instead dividing every term by n, you'd end up with:
\(\displaystyle \displaystyle \lim_{n \to \infty}{a_n}=\lim_{n \to \infty}{\frac{n^{1/2} + n^{1/3}}{n+2n^{2/3}}}=\lim_{n \to \infty}{\frac{n^{-1/2}+n^{1/3-1}}{1+2n^{2/3-1}}}=\lim_{n \to \infty}{\frac{\frac{1}{\sqrt{n}}+\frac{1}{n^{2/3}}}{1+\frac{2}{n^{1/3}}}}\)
From there, if you break it apart into the limits of the numerator and denominator, you get 0/1 = 0. So, you get the same answer no matter how you go about it. (Provided, of course, you do the math correctly
)
Assuming that the above is correct, then I agree with all of your workings, and the answer is correct. The small bit of working not shown is that the limit of a fraction is equal to the limit of its numerator divided by the limit of its denominator, excepting cases of indeterminate forms. That is to say:
\(\displaystyle \displaystyle \lim_{n \to \infty}{\frac{1+n^{-1/6}}{n^{1/2}+2n^{1/6}}}=\frac{\lim_{n \to \infty}({1+n^{-1/6}})}{\lim_{n \to \infty}{(n^{1/2}+2n^{1/6})}}=\frac{1}{\infty}=0\)
As to your confusion about whether to divide by sqrt(n) or n, the general wisdom is, as you noted, to divide by the highest power in the denominator. But, as you've also demonstrated, this is not a hard and fast rule. You don't always have to follow it exactly, although sometimes it means extra work if you don't. If you'd begun by instead dividing every term by n, you'd end up with:
\(\displaystyle \displaystyle \lim_{n \to \infty}{a_n}=\lim_{n \to \infty}{\frac{n^{1/2} + n^{1/3}}{n+2n^{2/3}}}=\lim_{n \to \infty}{\frac{n^{-1/2}+n^{1/3-1}}{1+2n^{2/3-1}}}=\lim_{n \to \infty}{\frac{\frac{1}{\sqrt{n}}+\frac{1}{n^{2/3}}}{1+\frac{2}{n^{1/3}}}}\)
From there, if you break it apart into the limits of the numerator and denominator, you get 0/1 = 0. So, you get the same answer no matter how you go about it. (Provided, of course, you do the math correctly
)