Limits And Derivatives - 1 - Free MCQ Practice Test with solutions, JEE

We have:
limx→0 ln(1+ax)]/x=a
lim x→0 ln(1+ax)/ax=a
and then:
lim x→0 [ln(1+ax)-ln(1+bx)]/x
= a-b

On rationalizing,
 Lt x-->0 (1-Cosx)/[√(1+x) +1]
= lt x-->0 2sin²(x/2)[√(1+x)+1] /x
= lt x-->0 [2sin(x/2)/x].sinx/2 [√(1+x) +1 ]
= 2.1/2 .0[1+1]=0

f(0) = lim(x→0) x sin (1/x)
​We know ∀ x ∈ R,sin (1/x) ∈ [−1,1]
Hence, f(0) = lim(x→0) xsin(1/x) 
= 0

We know that,

lim_x→3 (xⁿ – 3ⁿ)/(x – 3) = n(3)^n-1

Thus, n(3)^n-1 = 108 {from the given}

n(3)^n-1 = 4(27) = 4(3³) = 4(3)^4-1 = 108

Therefore, n = 4

Consider the given function.
lim x→0 ((1-x)ⁿ −1)/x)
This is 0/0 form.
So, apply L-Hospital rule,
lim x→0 [nx-1x(1-x)_^n-1 −0)]/1
lim x→ 0(-n(1-x)^n-1)
=-n

lim(x → 0) (tanx-x)/x² tanx
As we know that tan x = sinx/cosx
lim(x → 0) (sinx/cosx - x)/x²(sinx/cosx)
lim(x → 0) (sinx - xcosx)/(x² sinx)
lim(x → 0) cosx - (-xsinx + cosx)/(x²cosx + sinx (2x))
lim(x → 0) (cosx + xsinx - cosx)/x²cosx + 2xsinx)
lim(x → 0) sinx/(xcosx + 2sinx)
Hence it is 0/0 form, apply L hospital rule
lim(x → 0) cosx/(-xsinx + cosx + 2cosx)
⇒ 1/(0+1+2)
= 1/3

 lim x-->0 ((x^2cosx-)/1-cosx)
Differentiating numerator and denominator
lim x-->0 (2xcosx - x^2sinx)/sinx
Differentiating it for a second time
lim x-->0 (2cos - 2xsinx - 2xsinx - x^2cosx)/cosx
Substituting x = 0
= lim x-->0  2/1
= 2

Suppose that the Reqd. Limit L = lim x→0 (sinx − x)/x³.
Substiture x = 3y , so that, as x→0 , y→0.
∴ L = lim y→0 sin3y − 3y/(3y)³,
= lim y→0 (3siny − 4sin3y)− 3y)/27y³,
= lim y→0 {3(siny − y)/27y³) − (4sin³y/27y³)},
⇒ L = lim y→0 1/9 * (siny − y)/y³) − 4/27* (siny/y)³...(∗)
.Note that, here,
= lim y→0(siny − y)/y³)
= lim x→0 (sinx−x)/x³) = L.
Therefore, (∗) ⇒ L = 1/9*L − 4/27
or,  8/9L = −4/27
Hence, L = −4/27*9/8
=−1/6

f(x) - f(y) ≥ ln x - ln y + x - y
f(x) - f(y) ≥ (ln x - ln y) / (x - y) + 1
Let x > y
lim (as y → x) f'(x) ≥ 1/x + 1 .... (1)
Let x < y
lim (as y → x) f'(x) ≤ 1/x + 1 .... (2)
f'(x) = f'(x⁺)
f'(x) = 1/x + 1
f'(1/x) = x² + 1
∑ from n = 1 to 20 of (x² + 1) = ∑ from x = 1 to 20 of x² + 20
= (20 × 21 × 41) / 6 + 20
= 2890

Given:
f(x) = x³ + x² f'(1) + x f''(2) + f'''(3)
Taking derivatives:
f'(x) = 3x² + 2x f'(1) + f''(2)
f''(x) = 6x + 2f'(1)f'''(x) = 6
Substitute the values for f'(1), f''(2), and f'''(3):
f'(1) = -5, f''(2) = 2, f'''(3) = 6
So:
f(x) = x³ + x²(-5) + x(2) + 6
f'(x) = 3x² - 10x + 2
Now, substitute x = 10:
f'(10) = 3(10)² - 10(10) + 2
= 300 - 100 + 2
= 202
Hence, f'(10) = 202.

At x=0:

Since the numerator contains tan0 = 0, the whole numerator becomes 0, while the denominator is 1.

f(0)=0

lim(h → 0) [sin²(x+h) - sin²x]/h
l = lim(h→0) = [2sin(x+h)cos(x+h)]/1
{As we know that sin2x = 2sinxcosx
= sin2x

y = logₑ( (1 - x²) / (1 + x²) )

dy/dx = y' = -4x / (1 - x⁴)

Again,

d²y/dx² = y'' = -4(1 + 3x⁴) / (1 - x⁴)²

Again,

y' - y'' = -4x / (1 - x⁴) + 4(1 + 3x⁴) / (1 - x⁴)²

At x = 1/2,

y' - y'' = 736 / 225

Thus 225(y' - y'') = 225 × 736 / 225 = 736.

⇒ β + 3 = 0, α - 1 = 0 and - 1/2 - β/2 = 1/3
⇒ β = -3, α = 1
⇒ 2α - β = 2 + 3 = 5

P= lim (n → ∞) { (2^1/2 - 2^1/3) (2^1/2 - 2^1/5) ... (2^1/2 - 2^1/(2n+1))}
Let
2^1/2 - 2^1/3 → Smallest
2^1/2 - 2^(1/(2n+1)) → Largest
Using the Sandwich Theorem:
(2^1/2 - 2^1/3)ⁿ ≤ P ≤ (2^1/2 - 2(^1/(2n+1)))ⁿ
Since (value between 0 and 1)ⁿ → 0 as n → ∞, we get:
lim (n → ∞) (2^1/2- 2^1/3)ⁿ = 0
lim (n → ∞) (2^1/2 - 2^(1/(2n+1)))ⁿ = 0
∴ P = 0

Given limit can be written as

(By using expansion of sinx and tan x) ⇒ n²(a − n) − n = 0 ⇒ an − n² − 1 = 0

√x / √(16+√x) - 4

= (√x * √(16+√x) + 4) / ( √(16+√x) - 4) * (√(16+√x) + 4)

= (√x * √(16+√x) + 4) / 16+√x - 16

= √(16+√x) + 4

now at lim x->0

= √16 + 4 = 8

ax² + bx + 1 = a(x - α)(x - β)
αβ = 1/a
x² + bx + a = a(1 - αx)(1 - βx)
lim (x → 1/α) { (1 - cos(x² + bx + a)) / (2(1 - ax)²) } = lim (x → 1/α) { (1 - cos(a(1 - αx)(1 - βx))) / 2a²(1 -
αx)²(1 - βx)² }^(1/2)
= [ (1/2) * a² * (1 - β/α) ]^1/2
= (1/2) * 1/aβ * (1 - β/α) = 1/2 * 1/αβ * (1 - β/α)
= 1/2α * (1/β - 1/α) = 1/k * (1/β - 1/α)
Thus, k = 2α Ans.

By using the rule,

We get, req. limit