How do you find the limit of #cscx-cotx+cosx# as x approaches 0?

Question

Charles Arocha · Accepted Answer

Please see below.

#lim_(xrarro)cosx = 1#, so that term is not a problem, it's the other two that challenge us. In trigonometry, when in doubt, one thing to try is rewriting using just sines and cosines. #cscx-cotx = 1/sinx-cosx/sinx = (1-cosx)/sinx# I assume that we have learned that #lim_(thetararr0)sintheta/theta = lim_(thetararr0)theta/sintheta = 1# #" "# and #" "# #lim_(xrarrtheta)(1-costheta)/theta = 0# So we'll multiply by #x/x# to get #(x(1-cosx))/(xsinx) = x/sinx * (1-cosx)/x# So #lim_(xrarr0)(cscx-cotx +cosx) = lim_(xrarr0)(x/sinx * (1-cosx)/x+cosx)# # = (1) * (0) + 1 = 1#

Cameron Alexander · Answer

undefined To find the limit of cscx - cotx + cosx as x approaches 0, we can use the properties of trigonometric functions and basic limit rules.

First, let's simplify the expression:
cscx - cotx + cosx = 1/sinx - cosx/sinx + cosx = (1 - cosx + cos^2x)/sinx

Next, we can factor out a common factor of (1 - cosx):
(1 - cosx + cos^2x)/sinx = (1 - cosx(1 - cosx))/sinx = (1 - cosx)/sinx

Now, we can rewrite (1 - cosx)/sinx as (1/sinx - cosx/sinx):
(1 - cosx)/sinx = 1/sinx - cosx/sinx = cscx - cotx

As x approaches 0, sinx approaches 0, and cscx and cotx approach infinity. Therefore, the limit of cscx - cotx + cosx as x approaches 0 is infinity.