Home » Ask & Discuss » Mathematics. » Differential Calculus « Back to Discussion

Differential Calculus

Cool goIITian

Joined:	7 Feb 2012
Post:	44

11 Feb 2012 08:03:38 IST

0 People liked this

303

evaluate limit

Mathematics

limx->infinity{lim n->infinity ( { [1^2 (sinx)^x] + [2^2 (sinx)^x] +......+[n^2 (sinx)^x]} / n^3)} [.] denotes greatest integer function

Share this article on:

Comments (5)

jagdish singh

Blazing goIITian

Joined: 19 Jan 2008

Posts: 809

16 Feb 2012 16:53:29 IST

Like 0 people liked this

$\hspace{-16}$Here We have to Calculate $\mathbf{\lim_{x\rightarrow 0}\lim_{n\rightarrow \infty}\sum_{r=1}^{n}\frac{\lfloor r^2(\sin x)^x\rfloor}{n^3}\;,}$\\\\\\ Where $\mathbf{\lfloor x \rfloor = }$ Floor Function OR Here Greatest Integer Function\\\\\\\\ Let $\mathbf{x>0\;\;,(\sin x)^x=a}$ and Use the face $\mathbf{x-1< \lfloor x \rfloor \leq x}$\\\\\\ $\mathbf{a-1< \lfloor a \rfloor\leq a\Leftrightarrow r^2.(a-1)< \lfloor r^2.a \rfloor \leq r.a}$\\\\\\ $\mathbf{\sum_{r=1}^{n}r^2.(a-1)<\sum_{r=1}^{n}\lfloor r^2.a \rfloor\leq \sum_{r=1}^{n}r.a}$\\\\\\ $\mathbf{\lim_{n\rightarrow \infty}\sum_{r=1}^{n}\frac{r^2.a}{n^3}-\lim_{n\rightarrow \infty}\sum_{r=1}^{n}\frac{1}{n^3}<\lim_{n\rightarrow \infty}\sum_{r=1}^{n}\frac{\lfloor r^2.a \rfloor}{n^3}\leq \lim_{n\rightarrow \infty}\sum_{r=1}^{n}\frac{r.a}{n^3}}$\\\\\\ Using Reinman Sum, We Get like This.....\\\\\\ Let $\mathbf{\frac{r}{n}=x\Leftrightarrow \frac{1}{n}=dx}$ and Limits are $\mathbf{0\;,1}$\\\\\\$

$\hspace{-16}$So $\mathbf{a.\int_{0}^{1}x^2dx-0<\lim_{n\rightarrow \infty}\sum_{r=1}^{n}\frac{\lfloor r^2.a \rfloor}{n^3}\leq \int_{0}^{1}x^2dx}$ \\\\\\$\mathbf{\lim_{x\rightarrow 0}a.\frac{1}{3}<\lim_{x\rightarrow 0}\lim_{n\rightarrow \infty}\sum_{r=1}^{n}\frac{\lfloor r^2.a \rfloor}{n^3}\leq \lim_{x\rightarrow 0}\frac{1}{3}}$\\\\\\ Now Calculate $\mathbf{\lim_{x\rightarrow 0}a=\lim_{x\rightarrow 0}(\sin x)^x=1}$\\\\\\ Easily Calculate $\mathbf{0^0}$ form....\\\\\\ So $\mathbf{\lim_{x\rightarrow 0}\frac{1}{3}<\lim_{x\rightarrow 0}\lim_{n\rightarrow \infty}\sum_{r=1}^{n}\frac{\lfloor r^2.a \rfloor}{n^3}\leq \lim_{x\rightarrow 0}\frac{1}{3}}$\\\\\\ So Using SandWitch Theorem, We Get\\\\\\ $\boxed{\boxed{\mathbf{\lim_{x\rightarrow 0}\lim_{n\rightarrow \infty}\sum_{r=1}^{n}\frac{\lfloor r^2.a \rfloor}{n^3}=\frac{1}{3}}}}$$

Vishal U. Karve

Hot goIITian

Joined: 18 Jan 2012

Posts: 147

16 Feb 2012 17:20:58 IST

Like 0 people liked this

The question says lim x->infinity.

So, that guy atul posted the question incorrectly....it should have been lim x->0.

atul

Cool goIITian

Joined: 7 Feb 2012

Posts: 44

17 Feb 2012 17:48:28 IST

Like 0 people liked this

sin(x) will always fluctuate between -1,1.It will always be a fraction.Even if it approaches 1, it will approach it from only one side. i.e. it will always be<=1 not >=1.So the limit will be a fraction raised to infinity = 0this is your's statement and question is correct printed in arihant differential calculus page no. 300so finally question is correct either your previous proof is wrong if not then answer of this question is zero

Vishal U. Karve

Hot goIITian

Joined: 18 Jan 2012

Posts: 147

20 Feb 2012 18:39:20 IST

Like 0 people liked this

Jagdish bhaiya's solution uses x->0. That's why I said x->infinity may be wrong.

And you should know that Arihant is not free from errors.

My previous statement is correct that (sin x)^x tends to zero as x->infinity.

Vishal U. Karve

Hot goIITian

Joined: 18 Jan 2012

Posts: 147

20 Feb 2012 18:41:24 IST

Like 0 people liked this

See graph of (sinx)^x for validity of my statement.

Google this: (I mean copy and paste this in google)

y=(sin(x))^x

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team