Limit questions

Question

Evaluate the limit..

Lim _Δx-->0   sin ( x + Δx ) - sin x / Δx

 

Answer 1

By the definition of derivatives, you get the following

lim_Δx→0 [ sin( x + Δx ) - sin x ] / Δx = d/dx sin x = cos x, assuming that the angle x is measured in radians.

If you want to derive it, you will need to use the angle sum formula which is

sin(a+b) = sin a cos b + cos a sin b.

You get

lim_Δx→0 [ sin( x + Δx ) - sin x ] / Δx

= lim_Δx→0 [ sin x cos Δx + sin Δx cos x - sin x ] / Δx

= lim_Δx→0 (sin x)(cos Δ - 1) / Δx + lim _Δx→0 (cos x)(sin Δx) / Δx

= (sin x) lim_Δx→0 (cos Δx - 1) / Δx + (cos x) lim _Δx→0 (sin Δx) / Δx

Now all we need is to calculate the two limits

lim_Δx→0 (sin Δx) / Δx can be computed by drawing a picture. I won't draw it here, but from the correct picture you get that

sin Δx < Δx < tan Δx for 0 < Δx < Π/2

and since sin Δx, Δx, and tan Δx are odd functions,

sin Δx > Δx > tan Δx for -Π/2 < Δx < 0.

Dividing either inequality by Δx and taking reciprocal gives.

1 > (sin Δx) / Δx > cos Δx for -Π/2 < Δx < Π/2 and Δx ≠ 0.

Since lim_Δx→0 cos Δx = 1, the squeezing theorem implies that

lim_Δx→0 (sin Δx) / Δx = 1.

We can use this limit to evaluate the other limit as follows.

lim_Δx→0 (cos Δx - 1) / Δx

= lim_Δx→0 (cos Δx - 1)(cos Δx + 1) / [Δx (cos Δx + 1)]

= lim_Δx→0 (cos² Δx - 1) / [Δx (cos Δx + 1)]

= lim_Δx→0 (-sin² Δx) / [Δx (cos Δx + 1)]

= -(lim_Δx→0 (sin Δx) / Δx)(lim_Δx→0 (sin Δx) / (cos Δx + 1))

= 1*(sin 0) / (cos 0 + 1)

= 0 / 2

= 0

Plugging these limits in gives

lim_Δx→0 [ sin( x + Δx ) - sin x ] / Δx = 0 sin x + 1 cos x = cos x.