1ques 1):-limx->01/(sinx/x)(tanx/x)
1/limx->o(sinx/x)(tanx/x)
=1/1
=1
1) limx→0x2sinx.tanx
2) limx→0[x2sinx.tanx]
3) limx→0{x2sinx.tanx}
4)Find a,b
i) limx→∞(x2+1x+1-ax-b)=0
ii) limx→∞(√x2-x+1-ax-b)≥0
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8 Answers
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11Ans 4) (i) lim x→∞ (x2 +1x+1-ax-b) = 0
= lim x→∞ x2+1-ax2-ax-bx-bx+1 = 0
= lim x→∞ x2(1-a) - x(a+b) + (1-b)x+1 = 0
since the limit of expression is zero, therefore, degree of numerator < degree of denominator
So, numerator must be const , i.e, is a zero degree polynomial
Therefore, 1-a = 0 and a+b = 0
Thus, a=1 , b=-1
(ii) We have, lim x→∞ (√(x2-x+1) -ax-b ) ≥0
= lim x→∞ [x(1 - 1/x + 1/x2) 1/2 + ax-b] ≥ 0
= lim x→∞ [x{1 - 1/2(1/x - 1/x2) +.....} + ax-b] ≥ 0
= lim x→∞ [x(1+a) - (1/2 + b) - 1/2y +.......] ≥ 0
Therefore, 1+a ≥0 and (1/2 + b)≥ 0
Thus, a≥-1 and b≥ -1/2
24thx tush...becoz in my key ans were diff and obviously wrong [2][2]
112) sinx.tanx>x2 as cosx≈1 when x≈0, cos x actually temains less than 1....
24@tush ...thats the same soln i had...but what my real ques is ...
whats the inspiration for taking f(x) = sinx tanx - x2??? and doing all that diff work..
11Bcoz we have to examine the behaviour or tendency of the given function in the neighbourhood of x=0.