Test: Limits of Trigonometric Functions - Grade 9 MCQ

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

If we substitute the value of x=0 in numerator and denominator, we get the indeterminate form 0/0 . We should use L'hopital's rule.
lim x→ 0 (√1+2x-√1-2x)/sinx
Differentiate it we get
lim x→ 0 (1/√1+2x) + (1/√1-2x)/cosx
= (1/√1+2(0)) + (1/√1-2(0))/cos 0
=2

lim xsecx    x=0
lim x(1/cosx) = lim (x/cosx)
Put x = 0
= 0/cos0
= 0/1 = 0