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I have this limit, and i have no idea of approach: $$\lim_{n \rightarrow + \infty } \left(\frac{n^3}{4n-7}\right)\left(\cos\left(\frac1n\right)-1\right)$$ turns out to be of indeterminate form, how to solve it?

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As $n \to\infty$ we get a $"0\cdot\infty"$ kind of a limit to calculate. This can be done using L'Hopital's rule. – Amihai Zivan May 18 '12 at 10:58
Can't apply l'Hospital's rule because is limit of succession!!! If i solve it with Derive i have -1/8 result – Carmine Paternoster May 18 '12 at 11:11
And $-\frac18$ is correct. – Brian M. Scott May 18 '12 at 11:18
up vote 12 down vote accepted


let $x=\frac{1}{n}$


Applying L'hospital rule


Applying L'hospital rule again

$$\lim_{x\to0}\;\frac{-\cos(x)}{8-42x} =\frac{-1}{8}$$

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Thanks very much for your clear explanation – Carmine Paternoster May 18 '12 at 11:29

When you have an $\infty\cdot0$ indeterminate form, the standard trick is to convert it to an $\frac{\infty}{\infty}$ or $\frac00$ form by shifting one of the factors into the denominator. Here, for instance, you might try rewriting the limit as


since $\frac1{\cos(1/n)-1}$ doesn’t look like a very nice thing to have in your denominator. This is a genuine $\frac00$ form, so l’Hospital’s rule applies. (You may have to apply it more than once.)

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Can't apply l'Hospital's rule because is limit of succession!!! – Carmine Paternoster May 18 '12 at 11:10
@Carmine: You certainly can: it’s a basic fact about sequences when they’re defined by nice differentiable functions. – Brian M. Scott May 18 '12 at 11:11

$\cos(x) -1 = -2\sin^2(x/2)$ use it

let $x=\frac{1}{n}$, $\lim_{x\to0}\;\frac{-2\sin^2(x/2)}{4x^2-7x^3}$ = $\lim_{x\to0}\;\frac{-\sin^2(x/2)}{x^2/4}.\frac{1}{2.(4-7x)}$ =$-\lim_{x\to0}\;(\frac{\sin(x/2)}{x/2})^2 .\lim_{x\to0}\;\frac{1}{2.(4-7x)} $= $\frac{-1}{8}$

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$$\lim_{n\to\infty } \frac{n^3}{4n-7} \cos\left(\frac1n-1\right) = \lim_{n\to\infty} \frac{n}{4n-7} \frac{\cos\left(\frac1n-1\right)}{\frac1{n^2}}$$

Maybe you have memorized that $\lim\limits_{x\to 0} \frac{1-\cos x}{x^2}=\frac12$, if not, you can get this limits by applying L'Hospital rule twice. For $n\to\infty$ you have $\frac1n\to 0$, hence $$\lim\limits_{n\to\infty} \frac{\cos\left(\frac1n-1\right)}{\frac1{n^2}} = -\frac12.$$

The other limit $\lim\limits_{n\to\infty} \frac{n}{4n-7}$ should be easy.

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Hint $ $ The proofs by l’Hospital and power series essentially use (second) derivatives. One can eliminate these advanced techniques and use only derivatives. Changing variables $\rm\: x = 1/n\:$

$$\rm \lim_{x\to 0}\: \frac{cos(x)-1}{x^2}\:\frac{1}{4 - 7\:\!x}$$

has latter fraction $\to \dfrac{1}4\:$ and former $\rm\to \dfrac{cos''(0)}2 = -\dfrac{cos(0)}2 = -\dfrac{1}2\:$ by the formula

$$\rm f''(0) =\: \lim_{x\to 0}\:\frac{f(x) - 2\: f(0) + f(-x)}{x^2}\: \left[\:=\ 2\:\lim_{x\to 0}\frac{f(x)-f(0)}{x^2}\ \ if\ \ f(-x) = f(x) \right] $$

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If you have some background knowledge of power series, the following approach may be interesting. It uses the fact that a power series $\sum a_n(x-b)^n$ for $f(x)$ typically gives a very good indication of the behaviour of $f(x)$ near $x=b$.

It is convenient, but not necessary, to let $x=1/n$. Recall that the MacLaurin series for $\cos x$ is given by $$1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots.$$ Thus $$\cos x-1=-\frac{x^2}{2!}+O(x^4).$$ Note that $$\frac{n^3}{4n-7}=\frac{1}{x^2}\frac{1}{4-7x}.$$ So our product is $$\frac{1}{4-7x}\left(-\frac{1}{2!}+O(x^2)\right).$$ Finally, let $x\to 0^+$. The term $\frac{1}{4-7x}$ approaches $\frac{1}{4}$ and the $O(x^2)$ term approaches $0$.

Added: We can also use the following early calculus idea, which is in effect not far from the method used by Prasad G. We are interested in $$\lim_{x\to 0}\frac{\cos x -1}{4x^2-7x^3}.$$ Multiply top and bottom by $\cos x+1$, noting that then the top becomes $\cos^2 x-1$, which is $-\sin^2 x$. So we want
$$\lim_{x\to 0}\:\left(\frac{\sin x}{x}\right)^2\frac{-1}{(4-7x)(\cos x+1)}.$$ Let $x\to 0$, and use the fact that $\frac{\sin x}{x}\to 0$ as $x\to 0$.

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