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cos+ ?

Edit :

(1)

lim cos(x + 3π/2)/x = 0/0

x -> 0

0/0 ... use H ... -sin(x + 3π/2)/1 = -(-1)/1 = 1

(2)

lim (cosx +3π/2)/x = (1 + 3π/2) / 0 = +inf

x -> 0

lim cos(x+3π/2) / x as x→0

by putting x→0 we will have a form of 0/0

which requires the use of L'Hopital's rule:

-sin(x+3π/2)*1 / 1 =>-sin(x+3π/2)

-sin(3π/2) = 1

the limit is 1

Now that's interesting! What you prefer to do is to multiply this expression with (a million - cos x) / (a million - cos x), and rearrange. Doing that, we've lim (as x techniques 0) {sin(a million - cos x) / x * (a million - cos x) / (a million - cos x)} = lim (as x techniques 0) {sin(a million - cos x) / (a million - cos x) * (a million - cos x) / x} = lim (as x techniques 0) {sin(a million - cos x) / (a million - cos x)} * lim (as x techniques 0) {(a million - cos x) / x} Now, we've 2 usual theorems on indeterminate limits utilising sines and cosines. they're a million) lim (as x techniques 0) {sin x / x} = a million 2) lim (as x techniques 0) {(a million - cos x) / x} = 0 for this reason, we've lim (as x techniques 0) {sin(a million - cos x) / (a million - cos x)} * lim (as x techniques 0) {(a million - cos x) / x} = lim (as x techniques 0) {sin(a million - cos x) / (a million - cos x)} * 0 Now, we ought to make positive that the different decrease gained't bring about an indeterminate kind. we can make some substitutions, to make it extra reachable to be sure. enable A = a million - cos x As x techniques 0, A = a million - cos 0 A = a million - a million A = 0 A techniques 0. for this reason, substituting, we've lim (as x techniques 0) {sin(a million - cos x) / (a million - cos x)} * 0 = lim (as A techniques 0) {sin A / A} * 0 = a million * 0 = 0 for this reason, the decrease is 0. desire this helps!!!!!

I will also assume that you meant to type cos(x+ 3π/2) / x.

Note that cos(x + 3π/2) = cos(x) cos(3π/2) - sin(x) sin(3π/2) = sin(x).

It is a known result that sin(x)/x --> 1 as x-->0, so the answer is 1.

cos (x + 3π/2)/x as x --> 0 is an indeterminate form 0/0

thus you can apply L'Hopital rule

- sin(x + 3π/2) --> +1 (as x --> 0)