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Calculate the limit: $\lim\limits_{n\to\infty}n^2\left(\left(1+\dfrac{1}{n}\right)^8-\left(1+\dfrac{2}{n}\right)^4\right)$
My first suggestion was that $\lim\limits_{n\to\infty} = 0$. As in both brackets as ${n\to\infty}$: $\dfrac{1}{n}$ and $\dfrac{2}{n}$ will ${\to0}$, so it was going to be $\lim\limits_{n\to\infty}n^2\left(\left(1\right)^8-\left(1\right)^4\right) => \lim\limits_{n\to\infty} = 0$.
My second suggestion was using the properties of $\lim\limits_{n\to\infty}\left(1+\dfrac{1}{n}\right)^n = e$ and $\lim\limits_{n\to\infty}\left(1+\dfrac{k}{n}\right)^n = e^k$ to find limits of enclosing brackets expressions:
1. $\lim\limits_{n\to\infty}(1+\dfrac{1}{n})^8 => \lim\limits_{n\to\infty}((1+\dfrac{1}{n})^n)^{\frac{8}{n}} => e^{\frac{8}{n}}$
2. $\lim\limits_{n\to\infty}(1+\dfrac{2}{n})^4 => \lim\limits_{n\to\infty}((1+\dfrac{2}{n})^n)^{\frac{4}{n}} => (e^2)^{\frac{4}{n}} => e^{\frac{8}{n}}$
It brought me again to $\lim\limits_{n\to\infty}n^2 *(e^{\frac{8}{n}} - e^{\frac{8}{n}}) => \lim\limits_{n\to\infty} = 0$
However, $0$ is a wrong answer. How to find the limit?

P.S. I am self-study calculus newbie, so please answer as easy as possible (don't know L'Hôpital's rule yet).

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    $\begingroup$ Use binomial theorem. Terms upto $1/n^2$ matter. $\endgroup$ – Paramanand Singh Apr 5 '18 at 5:00
  • $\begingroup$ You can't just selectively evaluate limits like that $\endgroup$ – Andrew Li Apr 5 '18 at 5:09
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You have the product of two terms, one of which goes to $\infty$ and one of which goes to $0$. You cannot conclude that the product goes to $0.$

As to how to do the problem, you can expand out $\left(1+\frac{1}{n}\right)^8$ and $\left(1+\frac{1}{2n}\right)^4$ by the binomial theorem, multiply by $n^2$ and compute the limit.

The trick is that you don't have to go past the second degree terms, because once you have a power of $3$ or greater in the denominator, the product will go to $0$. For example, $$ \left(1+\frac{1}{n}\right)^8=1+\frac{8}{n}+\frac{28}{n^2}+o\left(\frac{1}{n^2}\right)$$ where by $o\left(\frac{1}{n^2}\right)$ I just mean terms that are very small compared to $1/n^2.$ Do the same thing with $\left(1+\frac{1}{2n}\right)^4$

Can you take it from here?

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  • $\begingroup$ yes, I managed to solve the limit. Could you please explain, why did you decide to multiply by n^2? Why not n, n^3, etc? $\endgroup$ – trthhrtz Apr 7 '18 at 5:05
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    $\begingroup$ Because of the factor of $n^2$ at the beginning. You can obviously do the problem by multiplying out the polynomials. The only trick is thinking about what they will entail, and seeing what part of the work is actually unnecessary. $\endgroup$ – saulspatz Apr 7 '18 at 5:10
  • $\begingroup$ Thanks for clarification. Now I get it where n^2 is coming from. I was thinking that you picked random variable :) $\endgroup$ – trthhrtz Apr 7 '18 at 5:15
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Let $x=1+\frac{1}{n}$ and $y=1+\frac{2}{n}$ then $$x^8-y^4=(x^2-y)(x^2+y)(x^4+y^2)$$ and

$$x^2-y=\frac{1}{n^2}$$

Thus the limit is $$(x^2+y)(x^4+y^2) \to 4$$

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  • $\begingroup$ Could you please explain, why did you imply that the last sum product goes to 4? $\endgroup$ – trthhrtz Apr 7 '18 at 5:06
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    $\begingroup$ Because $x\to 1$ and $y\to 1$ so $(x^2+y)(x^4+y^2)\to (1+1)(1+1)=4$ $\endgroup$ – Rene Schipperus Apr 7 '18 at 13:08
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Set $1/n=h$ to get $$\lim_{h\to0}\dfrac{(1+h)^8-(1+2h)^4}{h^2}$$

Now $\displaystyle(1+2h)^4=1+\binom41(2h)+\binom42(2h)^2+O(h^3)$

and $\displaystyle(1+h)^8=1+8h+\binom82h^2+O(h^3)$

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