JEE Advanced Level Test: Limit & Derivatives- 3 - JEE MCQ

 If lim(x → a) f(x) = lim(x → a) g(x) = 0
then, lim(x → a) [1 + f(x)]^1/g(x) = e^{lim(x → a) f(x)/g(x)}
= lim(x → α) (1 + ax^2 + bx + c)
= e^{lt(x → α) [a(x-α) (x-β)]/(x-α)} 
= e^[a(x-β)]

lim(x→∞) (f(x) + ((3f(x) − 1)/f²(x)) = 3
Let lim(x→∞) f(x) = y
or lim (x→∞) f(x) + [3lim(x→∞)f(x) − 1)]/(lim(x→∞) f(x))² = 3
or (y + (3y−1)/y²) = 3
or y³ − 3y² + 3y − 1 = 0
or (y−1)³ = 0
or y = 1

As α is root of ax² + bx + c = 0
 ∴ aα² + bα + c = 0. 
Now,

 

S = 1n + 2(n-1) + 3(n-2) +.................n(1)
S = (k → 1 to n)  k(n-k+1) = (k-->1 to n) (n+1)k - k²
= (n+1)[n(n+1)/2 - n(n+1)(2n+1)/6]
= n(n+1)/2 [(n+1) - (2n+1)/3]
= n(n+1)/2 * (n+2)/3 
= [n(n+1)(n+2)]/6
S/[1² + 2² + .....n²] = {[n(n+1)(n+2)]/6}/{[n(n+1)(2n+1)]/6}
= (n+2)/(2n+1)
= 1/2(1 + 3/(2n+1))
= lim(n → ∞) 1/2(1+3/(2n+1))
= 1/2

Solving tanα.x + sinα.y = α and α cosecα.x + cosα.y = 1, we get
x = (αcosα−sinα)/(sinα−α)
and y = (α−xtanα)/sinα
​lim x(α→0) = lim(α→0) (cosα − αsinα − cosα)/(cosα−1)
​= lim (α→0) αsinα/(2sin²(α/2))
= lim (α→0) [4(α/2)²(sinα/α)]/[(sinα/2)²]/2 = 2 
lim y(α→0) = lim(α→α) (α−xtanα)/(sinα)
= lim(α→0) (α/sinα − x/cosα)
= 1 − 2
= − 1
Hence P = (2,−1)

The question asks us about,
lim x→a_m (x−a₁)/|x−a₁| * (x−a₂)/|x−a₂| * (x−a₃)/|x−a₃|.......* (x−an)/|x−an|
lim x→a_m (x−a_m)/|x−a_m| doesnot exist because
lim x→a_m- (x−a_m)/|x−a_m| = -1
lim x→a_m+ (x−a_m)/|x−a_m| = 1
Since, LHL is not equal to RHL , hence the limit does not exist.
Therefore, the limit of the entire function does not exist

Let a = min{x²+2x+3:x∈R} and b = lim(θ→0) (1−cosθ)/θ²
​Now, x²+2x+3 
= x² + 2x + 1 + 2
=(x+1)² + 2
∴ min{x² + 2x + 3 : x ∈ R} = min{(x+1)2 + 2 : x ∈ R} = 2 
Also, b = lim(θ→0) (1−cosθ)/θ²
lim(θ→0) (sinθ/2θ) = 1/2 ...... {lim (θ→0) {sinθ/θ = 1}} 
∑(r = 0 to n)​     a^r ⋅b^n-r
= ∑(r = 0 to n) 2^r(1/2)^n-r
= 2^-n ∑(r = 0 to n) 2^2r
= 2^-n{1 + 2² + 2⁴ +.....2²ⁿ}
= 2-n . [1(4n+1 − 1)]/(4−1)
= [4ⁿ⁺¹ − 1]/[3.2ⁿ]