Test: Infinite Limits - Commerce MCQ

lim (x → 0) [((1-3x)+5x)/(1-3x)]^1/x
lim (x → 0) [1 + 5x/(1-3x)]^1/x
= e^{lim(x → 0) (1 + 5x/(1-3x) - 1) * (1/x)} 
= e^{lim(x → 0) (5x/(1-3x)) * (1/x)}
= e^{lim(x → 0) (5x/(1-3x))}
= e⁵

lim(x→0) [log10 + log1/10]/x
= [log10 + log10]/0
= 0/0 form
lim(x→0) [(1/(x+1/10) * 1]/1
lim(x→0) [(1/(0+1/10) * 1]/1
= 1/(1/10) => 10

lim(x → 1) (log2 2x)1/^log₂^x
= lim(x →1) (log22 + log2x)1/^log₂^x
As we know that {log ab = log a + log b}
lim(x → 1) {1 + log₂x}1/^log₂^x
log₂x → 0
Put t = log₂x
lim(t → 0) {1 + t}¹/t
= e

lim(x → 0) (tanx/x)^{(1/x^2)}
= (1)∞
e^{lim(x → 0) (1/x2)(tanx/x - 1)}
= e^{lim(x → 0) ((tanx - x)/x3)}   .....(1)
lim(x → 0) ((tanx - x)/x³)
(0/0) form, Apply L hospital rule
lim(x → 0) [sec²x -1]/3x²
lim(x → 0) [tan²x/3x²]
= 1/3 lim(x → 0) [tan²x/x²]
= 1/3 * 1
= e^1/3

lim(x → ∞) [(x-2)/(x+3)]^2x
lim(x → ∞) [(x-2)/(x+3)]² * [(x-2)/(x+3)]^x

lim(x → a) [f(x) * g(x)] = lim(x → a) f(x) * lim(x → a) g(x)

lim(x → ∞) [(x-2)/(x+3)]² * lim(x → ∞) [(x-2)/(x+3)]^x
Lets evaluate the limits of both the functions separately,
lim(x → ∞) [(x-2)/(x+3)]²
= lim(x → ∞) [(x(1-2/x)/(1+3/x)]²
lim(x → ∞) [(1-2/x)/(1+3/x)]²
Apply infinity property,
= [(1-0)/(1-0)]²
= 1
Now, lim(x → ∞) [(x-2)/(x+3)]²
= lim(x → ∞) exln[(x-2)/(x+3)]
= lim(x → ∞) ln[(x-2)/(x+3)]/(1/x)
Apply L hospital rule
lim(x → ∞) d/dx[ln(x-2)/(x+3)]/[d/dx(1/x)]
= lim(x → ∞) {1/[(x-2)/(x+3)] * d/dx[(x-2)/(x+3)]}/(-1/x2)
= lim(x → ∞) {(x+3)/(x-2)[(x+3)d/dx(x-2) - (x-2)d/dx(x+3)]/(x+3)²}/(-1/x2)
= lim(x → ∞) {(x+3)/(x-2)[(x+3)-(x-2)]/(x+3)2}/(-1/x2)
= lim(x → ∞) {(x+3)/(x-2)[5/(x+3)²]}/(-1/x2)
= lim(x → ∞) [-5x2/(x+3)(x-2)]
= lim(x → ∞) [-5x2/(x2 + x - 6)]
Again apply L hospital rule,
= lim(x → ∞) [d/dx(-5x2)/(d/dx(x2 + x - 6))]
= lim(x → ∞) [-10x/(2x + 1)]
Again applying L hospital rule,
= lim(x → ∞) [d/dx(-10x)/(d/dx(2x + 1))]
=  lim(x → ∞) [-10/2] 
=  lim(x → ∞) [-5]
= -5
= lim(x → ∞) exln[(x-2)/(x+3)]² = e(-5)2
= 1/e10
= lim(x → ∞) [(x-2)/(x+3)](x+3) = 1/e¹⁰ (1)
= 1/e¹⁰ = e^-10