1r u sure n is multiplied in the sixth q....its much similar q lik this.....http://targetiit.com/iit-jee-forum/posts/some-more-q-12956.html....
1) f(x) = i) Sin[x] ; [x] ≠0
[x]
ii) 0 when ; [x] = 0
Find -- Lim (x→0) f(x).
2) Lim(x→∞) {(x+6)/(x+4)}x+4 = ?
3) If x>0, Lim(x→0) {(sinx)1/x +(1/x)sinx} = ?
4) Lim(x→ -∞) {x4 Sin(1/x) +x2}/{1+ lx3l } = ?
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20 Answers
1
21i m so sorry but whole thing is not comming on screen....BUT THIS IS A VERY LENGTHY METHOD AND NOT AN ELEGENT one I GUESS.....waiting for some elegent method
21use exp of cos x
cosx=1-x22+x44!......
\lim_{x\rightarrow 0}\frac{1-(1-x^{2}/2!+x^{4}/4!....)^{5}(1-4x^{2}/2!+64x^{4}/4!......)^{3}(1-9x^{2}/2!+81x^{4}/4!.......)^{3}}{x^{2}}
now apply Lhosp
\lim_{x\rightarrow 0}-(\frac{5(1-x^{2}/2!+x^{4}/4!....)^{4}(-2x/2! +4x^3/4!....)(1-4x^{2}/2!+64x^{4}/4!......)^{3}(1-9x^{2}/2!+81x^{4}/4!.......)^{3} + 3(1-4x^{2}/2!+64x^{4}/4!......)^{2}(-8x/2!+ 256x^3/4!.....) (1-x^{2}/2!+x^{4}/4!....)^{5}(1-9x^{2}/2!+81x^{4}/4!.......)^{3}+3(1-9x^2/2!+81x^4/4!......)^2(-18x/2!+324x^3/4!........)(1-x^{2}/2!+x^{4}/4!....)^{5}(1-4x^2/2!+64x^4/4!.......)^3 }{2x})
now cancelling x from numerator and denominator
\lim_{x\rightarrow 0}-(\frac{5(1-x^{2}/2!+x^{4}/4!....)^{4}(-2/2! +4x^2/4!....)(1-4x^{2}/2!+64x^{4}/4!......)^{3}(1-9x^{2}/2!+81x^{4}/4!.......)^{3} + 3(1-4x^{2}/2!+64x^{4}/4!......)^{2}(-8/2!+ 256x^2/4!.....) (1-x^{2}/2!+x^{4}/4!....)^{5}(1-9x^{2}/2!+81x^{4}/4!.......)^{3}+3(1-9x^2/2!+81x^4/4!......)^2(-18/2!+324x^2/4!........)(1-x^{2}/2!+x^{4}/4!....)^{5}(1-4x^2/2!+64x^4/4!.......)^3 }{2})
now putting limit of x as 0
we get
-(5(1)(-2/2!)(1)(1) +3(1)(-8/2!)(1)(1) +3(1)(-18/2!)(1)(1))2
=-(-5-12-27)/2
=-(-44)/2=22
13Pata nahi, i've done thrice this way, n my ans comes same each time...
So, either this approach is wrong, or am doing the same mistake over n over again..!![3]
1ok ok....u did right...in the 6th...but wen i calc it was coming at teh end something (n.something)...i mean n was not getting cancelled
13HeHee yes a typo again, editing...
n in 6) I expanded binomially ignoring the higher terms, can't tht be done ?
1in ur 7th q....r u sure u wrote the q corectly....i mean u wrote cos3x and cos5x seperatly.....without clubing them...i guess u made a typo....and even if its that wat u wrote my an not matching......u hav to jus use exp of cosx in it
13One more Limit to be evaluated-
7) Lim(x→0) \frac{1- cos^5x.cos^32x.cos^33x}{x^2}
Ans: 22
13Yup, tht "n" is there outside...
Nyways this is my sol.n, dunno where am i going wrong...
Taking out "n" common frm inside -
Lim(n→∞) n^2 \left[ \left(1+\frac{3}{n}+\frac{2}{n^2}+\frac{1}{n^3} \right)^{1/3} + \left(1-\frac{2}{n}+\frac{3}{n^2}\right)^{1/2} -2\right]
As n→∞, Expressions {λ}/n are very small, hence -
Lim(n→∞)n^2 \left[ \left(1+\frac{1}{n}+\frac{2}{3n^2}+\frac{1}{3n^3} \right) + \left(1-\frac{1}{n}+\frac{3}{2n^2}\right) -2\right]
Which finally reduces to-
LIm(n→∞) n^2 \left[ \frac{2}{3n^2}+\frac{1}{3n^3} +\frac{3}{2n^2} \right]
Taking n2 inside & Applying Limit, it wud finally give =>
2/3+3/2 = 13/6.
Plz point out my mistake, if any, kyunki answer is not matching...
62First question..
right limit is zero
Left hand limit is -( sin -1 ) = sin 1
hence the limit does not exist..
106Q5. using LH,
(10xln10 - 2xln2 -5xln5)/tanx
LH again
(10xln210 - 2xln22 -5xln25)/sec2x
= ln210 - ln22 - ln25
13Some More...
5) A doubt- Lim(x→0) (10x -2x -5x +1)/(ln(secx))
6)Lim(n→∞) n{3√n3+3n2+2n+1 + √n2-2n+3 - 2n}
Am getting 13/6 as the ans.
106Q4. as x--> -∞ , lxl = -x
So the limit is (x4sin(1/x) + x2)/(1-x3)
writing sin(1/x) as 1/x
we get lim(x--> -∞) (x3+x2)/(1-x3) = -1
106Q3. consider sinx1/x = y
lny = ln(sinx)/x as x-->0+ lnsinx --> -∞ and x--> 0+ so lny --> -∞
=>y -> e-∞ =0
consider (1/x)sinx =y
lny = sinx*ln(1/x) = ln(1/x)/(1/x) usnig LH.,
lny --> x = 0
=> y --> e0 = 1
so the limit is 0+1 = 1
13Sorry forgot 2 mention tht Bhaiyaa [3] ; Not my doubts, just fr practice...
@LionKing.... U making a silly mistake i think..
(P.S. Only In search of a perfect sol.n fr Q3, 'coz am a bit uncertain about my own, though i got the ans.. :p)