JEE Advanced Level Test: Limits & Derivatives- 1 - JEE MCQ

For left hand limit x-1 will tend to 0
However for greatest integer function (second term) since x is slightly lesser than 1, x-1 tends to left side of zero and that under G.I.F. will be equal to -1.
For the third term since x is lesser than 1, 1-x will tend to right side of 0 and that under the G.I.F. will be equal to zero.
So the final limit will be equal to 0–1+0 = -1

lim x-->∞ sec^-1(x/x+1)
lim x-->∞ sec^-1(x/x[1+1/x])
lim x-->∞ sec^-1(1/(1+1/x))
sec^-1(1/(1+1/∞))   {therefore 1/∞ = 0)
sec^-1 (1/(1+0))
sec^-1(1)
sec^-1(sec0)
= 0

We know from trigonometry that
−1≤sin(1x)<−1 for all x≠0
Important: for limx→0 we don't care what happens when x=0
Since x<2>0 for all x≠0 , we can multiply through by x² to get
−x²= x²sin(1/x)≤x²
Clearly limx→0(−x²)= 0 and limx→0 x²=0, so, by the squeeze theorem,
lim x→0 x²sin(1/x) = 0

cosx=sin(x−π/2) 
limx→π/2 (x−π/2)/cosx
limx→π/2 (x−π/2)/[sin(x−π/2)]
= limθ→0 [θ/sinθ]= 1 
we let  θ=x−π/2  part way through
since  limt→0 sint/t=1  is a given after the first chapter in most calculus classes.
Using L’H: 
limx→π/2 (x−π/2)/cosx
= limx→π/2  (1/(−sinx)) = -1

 lim x→π/2 tan²x(√2sin²x+3sinx+4 − √sin²x+6sinx+2)
lim x→π/2 tan²x[(√2sin^2x+3sinx+4 − √sin²x+6sinx+2) *  (√2sin²x+3sinx+4 +√sin²x+6sinx+2)]
a = 2sin²x+3sinx+4 
b = sin²x+6sinx+2
(a+b) * (a-b) = a² - b²
=lim x→ pi/2  tan²x((2sin²x+3sinx+4) − (sin²x+6sinx+2))/[√2sin²x+3sinx+4 + √sin²x+6sinx+2]
=lim x→ pi/2 tan²x[sin^2x+3sinx+2]/[√2sin²x+3sinx+4 + √sin²x+6sinx+2]
=lim x→ pi/2 tan^2x = sin²x/cos²x
tan²x = sin²x/(1-sin²x)
=lim x→ pi/2 -sin²x(sinx-1)(sinx-2)/(1-sinx)(1+sinx)
=lim x→ pi/2 -sin(pi/2) sin²(pi/2-x)/[1+sinpi/2) * 6
= -1*(-1)/2*6  = 1/12

lim θ→0[nsinθ/θ]=n−1.....[from graph maximum value of sinx/x=1]
lim θ→0[ntanθ/θ]=n.....[from graph minimum value of tanx/x=1]
Hence, lim x→θ([[nsinθ/θ]+[ntanθ/θ])
= n−1+n  =2n−1

Hence it is 0/0 form, Use L hospital rule,
Log y = log 2^-cosx
Log y = -cosx log2
1/y dy/dx = - log2[-sinx]
1/y dy/dx = sinx log2
dy/dx = ysinx log2
dy/dx = 2^-cosx log2 sinx
Lim x→ pi/2  (2^-cosx log2 sinx - 0)/((pi - pi/2)+x)
Lim x→ pi/2  (1*log2*1)/pi/2
= 2log2/pi 
=(log2²)/pi
= 2log2/pi

Apply L′Hopital′s Rule

Apply L′Hopital′s Rule