Test: Limits (16 August) - JEE 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

sin3x/sin4x = sin3x/(1)⋅(1/sin4x)
= [(3x/1)sin3x/3x] ⋅ [1/4x(4x/sin4x)]
= 3x/4x[sin(3x)/3x(4x/sin(4x))]
= 3/4 [sin3x/3x/(4x/sin4x)]
Now, as x→0, (3x)→0 so sin3x/3x→1.(Using θ= 3x)
And, as x→0, (4x)→0 so 4x/sin4x→1. (Using θ = 4x)
Therefore the limit is 3/4
lim x→0 sin3x/sin4x
= lim x→0   3/4sin3x/3x(4x/sin4x)
= 3/4 lim x→0 sin3x/3x
lim x→0 4x/sinx
= 3/4(1)(1)
= 3/4

lim (x-> 0) (1-cos2x) / x

= lim (x->0) {1-(1–2sin² x)}/x

= lim (x->0) 2sin^x/x

When we put the limit…the answer is 0/0 and it is undetermined.

So we have to apply LHR rule to determine it,

= lim(x->0) 2 * 2 sin x cos x / 1

= lim(x->0) 2 sin 2x

Now put the limit, we get

2 * sin 0 = 0