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QUESTION: 1

Solution:

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

QUESTION: 2

The value of

Solution:

QUESTION: 3

Solution:

QUESTION: 4

The value of the limit

Solution:

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

= lim (x->0) {1-(1–2sin^{2} 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

QUESTION: 5

The value of

Solution:

QUESTION: 6

Solution:

QUESTION: 7

Evaluate

Solution:

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

QUESTION: 8

The value of

Solution:

lim xsecx x=0

lim x(1/cosx) = lim (x/cosx)

Put x = 0

= 0/cos0

= 0/1 = 0

QUESTION: 9

Solution:

QUESTION: 10

Solution:

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