Help me evaluate limit of sequence 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?
 A: 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
$$\lim_{n\to\infty}\frac{\cos(1/n)-1}{\frac{4n-7}{n^3}}\;,$$
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.)
A: $\cos(x) -1 = -2\sin^2(x/2)$ 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)} $= $\frac{-1}{8}$    
A: $$\lim_{n\to\infty}\frac{\cos(1/n)-1}{\frac{4n-7}{n^3}}\;,$$
let $x=\frac{1}{n}$
$$\lim_{x\to0}\;\frac{\cos(x)-1}{4x^2-7x^3}$$
Applying L'hospital rule
$$\lim_{x\to0}\;\frac{-\sin(x)}{8x-21x^2}$$
Applying L'hospital rule again
$$\lim_{x\to0}\;\frac{-\cos(x)}{8-42x} =\frac{-1}{8}$$
A: $$\lim_{n\to\infty } \frac{n^3}{4n-7} \cos\left(\frac1n-1\right) = \lim_{n\to\infty} \frac{n}{4n-7} \frac{\cos\left(\frac1n-1\right)}{\frac1{n^2}}$$
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.
A: 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
$$\rm f''(0) =\: \lim_{x\to 0}\:\frac{f(x) - 2\: f(0) + f(-x)}{x^2}\: \left[\:=\ 2\:\lim_{x\to 0}\frac{f(x)-f(0)}{x^2}\ \ if\ \ f(-x) = f(x) \right] $$
A: 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 $x\to 0$. 
