Evaluate the limit x>0 of (xcotx)?

Question

can someone please explain how to get the answer step by step?? thanks a lot

Answer 1

We have cotx = cosx/sinx, , for sinx <>0. Therefore, x cotx = x cosx/sinx = (x/sinx) cosx. (1)

According to a well known fact, lim x -> 0 sinx/x =1. Since 1 <>0, the properties of limits of functions imply that lim x-> 0 x/sinx = 1/( lim x-> 0 x/sinx) = 1/1 = 1. In addition, we know the cosine function is continuous, so tha lim x-> 0 cosx = cos(0) = 1.

From (1), we see x cotx is given by the product of 2 functions that have a limit at x = 0. We know that, in such cases, the limit of the product is the product of the limits. Therefore,

lim ( x-> 0) x cot x = lim ( x-> 0) (x/sinx) cosx = lim (x -> 0) x/sinx) . lim (x -> 0) cos x = 1 . 1 = 1

Answer 2

thats untrue because cot is infinate. u can't evaluate the limit by substitution because cot 0 is undefined. you use l'hopitals rule i guess. its basically x/tanx right? so differentiate x (1) and tan x ( which is sec ^2 x) and then u can substitute.

Answer 3

xcot(x)
=x cos(x) / sin(x)
= (x/sin(x)) cos(x)
Limit of the product is the product of the limits.
Limit of x/sin(x) as x approaches 0 is 1. Cos(x)=1 for x=0.
Limit is therefore 1.

Answer 4

it is Zero
as x gets closer to zero x times anything equals zero.