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Find

$$\lim_{n \rightarrow \infty} \dfrac{1-\left(1-\dfrac{1}{n}\right)^4}{1-\left(1-\dfrac{1}{n}\right)^3}$$

I can't figure out why the limit is equal to $\dfrac{4}{3}$ because I take the limit of a quotient to be the quotient of their limits.

I'm taking that $\lim_{n \rightarrow \infty}1-\left(1-\frac{1}{n}\right)^4 = 0$ and likewise that $\lim_{n \rightarrow \infty}1-\left(1-\frac{1}{n}\right)^3 = 0$, which still gives me that the limit should be 0.

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    $\begingroup$ The rule your using only applies when the limit of the denominator is not 0. Remember that $0/0$ is undefined. $\endgroup$ – Tim Raczkowski Jan 13 '15 at 4:30
  • $\begingroup$ Okay thanks! Any rules I should apply first then because I'm still not seeing how each part could possibly be an integer after 1- anything. $\endgroup$ – Bee Jan 13 '15 at 4:32
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    $\begingroup$ See GabrielH's solution. He is using L'Hospital's rule. $\endgroup$ – Tim Raczkowski Jan 13 '15 at 4:35
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    $\begingroup$ If you are still early in your course and haven't established "LHR" yet (or are learning how to compute limits without using it -- as many questions on this site do), you may want to go with one of the other two solutions. $\endgroup$ – colormegone Jan 13 '15 at 5:36
  • $\begingroup$ @Arjang It is better not to use things like \dfrac in titles, see meta. $\endgroup$ – Martin Sleziak Jan 13 '15 at 9:00
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$$\lim_{n \rightarrow \infty} \dfrac{1-\left(1-\dfrac{1}{n}\right)^4}{1-\left(1-\dfrac{1}{n}\right)^3} \stackrel{\mathscr{L}}{=}\lim_{n \rightarrow \infty} \dfrac{4\left(1-\dfrac{1}{n}\right)^3 \dfrac{1}{n^2}}{3\left(1-\dfrac{1}{n}\right)^2\dfrac{1}{n^2}} =\lim_{n \rightarrow \infty} \dfrac{4\left(1-\dfrac{1}{n}\right)^3}{3\left(1-\dfrac{1}{n}\right)^2} = \color{blue}{\dfrac{4}{3}}$$

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If you expand the binomials, you get $\dfrac{1-\left(1-\dfrac{1}{n}\right)^4}{1-\left(1-\dfrac{1}{n}\right)^3}=\dfrac{1-\left(1-\dfrac{4}{n}+\dfrac{6}{n^2}-\dfrac{4}{n^3}+\dfrac{1}{n^4}\right)}{1-\left(1-\dfrac{3}{n}+\dfrac{3}{n^2}-\dfrac{1}{n^3}\right)}=\dfrac{\dfrac{4}{n}-\dfrac{6}{n^2}+\dfrac{4}{n^3}-\dfrac{1}{n^4}}{\dfrac{3}{n}-\dfrac{3}{n^2}+\dfrac{1}{n^3}}$. Multiplying by $n$ on the top and bottom yields $\dfrac{4-\dfrac{6}{n}+\dfrac{4}{n^2}-\dfrac{1}{n^3}}{3-\dfrac{3}{n}+\dfrac{1}{n^2}}$. Then the limit is easier to evaluate. Any term with a division by $n$ will drop out as $n\to \infty$.

$\lim_{n \to \infty} \dfrac{1-\left(1-\dfrac{1}{n}\right)^4}{1-\left(1-\dfrac{1}{n}\right)^3}=\dfrac{4-0+0-0}{3-0+0}=\dfrac{4}{3}$

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  • $\begingroup$ I took the liberty to edit the answer for better readability. See if it's fine. $\endgroup$ – Krish Jan 13 '15 at 6:21
  • $\begingroup$ Looks better to me. $\endgroup$ – Benjamin Roycraft Jan 13 '15 at 22:10
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Hint: $\dfrac{1-x^4}{1-x^3}= \dfrac{(1-x)(1 + x + x^2 + x^3))}{(1-x)(1+x+x^2)} = \dfrac{1 + x + x^2 + x^3}{1 + x+ x^2}$ (provided $x \neq 1$.)

Now put $x = 1 - \frac{1}{n}$ and take $n \to \infty.$

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  • $\begingroup$ It's going to take a lot of work to sort this out if you're putting $1-\frac{1}{n}$ into every $x$ position. $\endgroup$ – Benjamin Roycraft Jan 13 '15 at 4:45
  • $\begingroup$ @BenjaminRoycraft: not necessarily!! $x_n \to x \Rightarrow x_n^2 \to x^2, x_n^3 \to x^3.$ these are standard facts. but if you don't want to use these and want more elementary method, then yours answer is the best one. $\endgroup$ – Krish Jan 13 '15 at 4:50
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    $\begingroup$ Yes, you're right. For some reason, I thought your final form was still indeterminate. You can directly evaluate the limit from your final expression. $\endgroup$ – Benjamin Roycraft Jan 13 '15 at 4:53

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