Limits Season

use 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

no no..... tahts the way to do such type of Qs...no prob with that [1]

Pata 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]

One more Limit to be evaluated-

7) Lim(x→0) \frac{1- cos^5x.cos^32x.cos^33x}{x^2}

Ans: 22

Yup, 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...

Q5. using LH,

(10^xln10 - 2^xln2 -5^xln5)/tanx

LH again

(10^xln²10 - 2^xln²2 -5^xln²5)/sec²x

= ln²10 - ln²2 - ln²5

Q4. 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

Q3. consider sinx^1/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

Sorry 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)

2nd one is e² ....1^∞ form