1I hv solved them. just i want confirmation of answers as i don't hv right now.
Q.1) \lim_{x\rightarrow 0}(\left[f(x) \right]+x^2)^1^/^\left\{f(x) \right\}, where f(x) = (tanx/x)
and \left\{f(x) \right\} denotes fractional part of f(x) and [f(x)] denotes the greatest integer function of f(x). is equal to ?
Q.2) \lim_{n\rightarrow infinity}n^2(x^1^/^n -x^1^/^n^+^1) = ?
Q.3) \lim_{n\rightarrow infinity}\sum_{r=1}^{n}{(1/2^r)tan(x/2^r)} = ?
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15 Answers
1
213> 1x - cotx
same as finding the sum of
12tan x2 + 122tan x22............
u must hav done this one
19yes...thats right !
we use 1/2 tan x/2 + 1/4 tan x/4 + ......1/2n tan x/2n = 1/2n cot x/2n - cot x (knowing that cot θ - tanθ =2cot 2θ)
....finally we get the limit as 1/x - cotx
21for q2 i guess jus take t=1/n and then apply l hospital....tat will do i guess
1@tushar u made a mistake der
x is constant here[1]
so correct the differntiation while apply l hospital
1ya , i also got 1) e3 and 3) 1/x - cotx.
but, Q.2 ,plz provide the solution. i'll get the answers today.
12)
\lim_{n\rightarrow \infty} n\left(\frac{x^{1/n}-1}{1/n}-\frac{n}{n+1}\frac{x^{1/(n+1)}-1}{1/(n+1)} \right)
=\lim_{n\rightarrow \infty} n\left(lnx-\frac{n}{n+1}lnx \right)
=\lim_{n\rightarrow \infty} \frac{n\: lnx}{n+1}
=lnx
1great solution sir. i'm providing my solution.
[\lim_{h\rightarrow 0}(x^h-x^h^/^h^+^1)] / h2 , where h → 1/n.
now, applying L'HOSPITAL RULE
\lim_{h\rightarrow 0}(hx^h^-^1-[h/(h+1)]x^-^1^/^h^+^1)/ 2h
Now , cancelling h and putting limits, we get :
x-1 - x-1 / 2 = 0
i'm not able to find the mistake in it.
21x is a constant here...[1]
u made a mistake in differentiation
21but bhaiya don u think differntiation of xh/(h+1) wrt h will be
xh/(h+1) lnx d( h/(h+1))dh
differentiation of ax is ax lna
1my bad.... when i'll leave my habit of making silly mistakes :(
yes deepak u r correct.... deleted my above post to avoid confusions