Evaluate the following limit:
lim x->0 xcscx
=limx->0 x/sinx
Does it equal one because the reciprocal is 1/1=1?
good logic
To evaluate the limit of xcscx as x approaches 0, we can simplify the expression by using the identity cscx = 1/sinx.
lim x->0 xcscx
= lim x->0 x/(sinx)
Now, let's analyze the expression x/sinx as x approaches 0.
When x approaches 0, sinx also approaches 0. Therefore, we have an indeterminate form of 0/0.
To evaluate this indeterminate form, we can use L'Hôpital's rule, which states that if we have a limit of the form 0/0 or ∞/∞, we can take the derivative of the numerator and denominator separately and then evaluate the limit again.
Taking the derivative of the numerator x and denominator sinx using the quotient rule, we get:
lim x->0 d/dx (x) / d/dx (sinx)
= lim x->0 1/cosx
Now, we can substitute x = 0 in the expression:
lim x->0 1/cosx
= 1/cos(0)
= 1/1
= 1
Therefore, the limit of xcscx as x approaches 0 is indeed 1.