4

$$ \lim_{n\to\infty}\frac{(2n+2)(2n+1)}{(n+1)^{2}} $$

When I expand it gives: $$ \lim_{n\to\infty} \dfrac{4n^{2} + 6n + 2}{n^{2} + 2n + 1} $$ How can this equal $4$? Because if I replace $n$ with infinity it goes $\dfrac{\infty}{\infty}$ only.

N. F. Taussig
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Deepak
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5 Answers5

5

$$\lim_{n\rightarrow \infty }\frac{(2n+2)(2n+1)}{(n+1)^2}=\lim_{n\rightarrow \infty }\frac{(2n/n+2/n)(2n/n+1/n)}{(n/n+1/n)^2}=\lim_{n\rightarrow \infty }\frac{(2+2/n)(2+1/n)}{(1+1/n)^2}=\frac{(2+0)(2+0)}{(1+0)^2}=4$$

or $$\lim_{n\rightarrow \infty }\frac{(2n+2)(2n+1)}{(n+1)^2}=\lim_{n\rightarrow \infty }\frac{(2)(2n+1)}{(n+1)}=\lim_{n\rightarrow \infty }\frac{(2)(2n/n+1/n)}{(n/n+1/n)}=\frac{(2)(2+0)}{(1+0)}=4$$

E.H.E
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4

To handle limits involving fractions you factor out the dominant term from top and bottom such that they cancel, by dominant term; I mean the term of highest degree, in this case it's $n^2$: $$\lim_{n\to\infty} \frac{4n^2 + 6n + 2}{n^2 + 2n + 1}$$ $$=\require\cancel\lim_{n\to\infty} \frac{2\cancel{n^2}\left(2 + \frac{3}{n} + \frac{1}{n^2}\right)}{\cancel{n^2}\left(1 + \frac{2}{n} + \frac{1}{n^2}\right)}$$ $$=\lim_{n\to\infty} \frac{2\left(2 + \frac{3}{n} + \frac{1}{n^2}\right)}{1 + \frac{2}{n} + \frac{1}{n^2}}\tag{1}$$ $$=\frac{2\left(2 + 0 + 0\right)}{1 + 0 + 0}\tag{2}$$ $$=\color{blue}{4}$$

You get from $(1)$ to $(2)$ by making the observation that each of the fractions with $n$ or $n^2$ in the denominator will equal zero in the limit as $n$ tends to infinity.

BLAZE
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3

$$\lim_{n\rightarrow \infty }\frac{(2n+2)(2n+1)}{(n+1)^2}=\lim_{n\rightarrow \infty }\frac{2(2n+1)}{n+1}=2\lim_{n\rightarrow \infty }\frac{2n+2-1}{n+1} =2\left(2-\lim_{n\rightarrow \infty }\frac{1}{n+1}\right)=2\left(2-\lim_{n\rightarrow \infty }\frac{1/n}{1+1/n}\right)=2(2)=4$$

James Pak
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  • Nice method, (+1) I never would have thought to factor out the $2$ from the first bracket to get cancellation of $(n+1)$: well done! – BLAZE Oct 13 '15 at 10:43
  • That was basically my approach, too. :) But I think you're last step isn't necessary, since we can safely say that $\lim_{n\rightarrow \infty }\frac{1}{n+1}=0$. – PM 2Ring Oct 13 '15 at 11:37
2

HINT: rewrite it in the form $$\frac{4+\frac{6}{n}+\frac{2}{n^2}}{1+\frac{2}{n}+\frac{1}{n^2}}$$

Dr. Sonnhard Graubner
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1

With equivalents:

At $\infty$, a polynomial is equivalent to its highest degree term, hence a rational function (quotient of polynomials) is equivalent to the the ratio of the highest degree terms of its numerator and its denominator: $$\frac{(2n+2) (2n+1)}{(n+1)^2}\sim_\infty\frac{(2n)^2}{n^2}=4.$$

Bernard
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