Tricky limit problem, give it a try. Question below?

Question

lim ........ [(sin 4x)/(8x)] =
x=>0


I believe it is 4/5

Answer 1

lim (x->0) (sin 4x) / (8x) =
lim (x->0) (4/4) (sin 4x) / (8x) =
lim (x->0) (4/8) (sin 4x) / (4x) =
(1/2) lim (x->0) (sin 4x) / (4x) =
1/2,
because lim (x->0) (sin ax) / (ax) = 1, for any nonzero number a.

Answer 2

Because sin(4x)/(8x) = 0/0 for x=0, then you can apply L'Hopital's Rule, which says that the limit f(x)/g(x) = limit f'(x)/g'(x).

The derivative of sin(4x) with respect to x is 4*cos(4x).

The derivative of 8x with respect to x is 8.

So the limit you want to evaluate is the limit of

4*cos(4x)/8

as x->0. However, for x=0, cos(4x)=1. Thus, the result is 1/2.


Alternatively, consider that for x very small, sin(4x) ~ 4x. In fact, |sin(4x)|<=|4x|. Thus, sin(4x)/8x is going to look like 1/2 as x goes to 0.

Answer 3

lim sin4x/8x x=>0
=[1/2]limsin4x/4x x=>0
put 4x=z
=1/2*[lim sinz/z] z=>0 we know this =1
=1/2*1
=1/2

Answer 4

You can use Hopital rule: find derivation's of sin(4x) and 8x then:
lim(4cos(4x)/8)=4/8=1/2
x==>0

Answer 5

Surely the limit is infinity because as x tends to 0, so does 8x, which means that you are dividing by 0, which gives you infinity.

Answer 6

It is tricky. The answer is 1/2. Was this the trick?

Answer 7

HUH?