71I didn't get any idea on how to proceed with Sandwich theorem.
1. \lim_{x\rightarrow 0^{-}} \frac{\sum_{r =1}^{2n+1}{[x^{r}]+(n+1)}}{1+[x]+|x|+2x}
2. \lim_{x\rightarrow 0} \left\{\lim_{n\rightarrow \infty } \left(\frac{[1 (sinx)^{x}]+[2^{2}(sinx)^{x}] + .....}{n^{3}} \right) \right\}
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717133 ... (may be.. I tookk it down from a book by SK GOYAL (JPNP) .. so I noted first two and then ....)
1.PROBLEM-1: "I" "believe" that the answer would depend on n and the answer i m getting is "n"
tell me if it is right!
1for the second problem i feel we need one more term to understand what the problem setter wanted!
1it should be 32
an attempt.....taking 1st term,applying SQUEEZE THEOREM
12(sin x)x-1<[12(sin x)x]<=12(sin x)x
:
:
write similar tems till n,then add ,
Σr2(sin x)x-n <Σ[r2(sin x)x]<=Σr2(sin x)x
now divide by n3
and take Lim n→∞
you get the expression as Lim x→∞(1/3)(sin xx)
so ans is 1/3
1@samagra : how did u evaluate limx→0 (sinx)x ??
00 is defined ?? [135]
212nd one an informal solution will be to put (sinx)^x = 1 where x is an infinitesimal. Now the job is done. Answer is 1/3
21It cant be 3^3 , 4^4 .... Because this way the sequence will diverge..
1That way it is coming 1/3.....but my confusion didn't clear with the logic of the final statement.. [7]
211st one the numerator is exactly zero. The denominator is almost zero. So the required limit is zero.
1@abhishek >>FOR YOU
take log and apply limit
x log sinx=(log sinx)/(1/x), (which is in ∞ /∞ form)
now apply LH rule
you get limit as 0
so e0 =1