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

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Answer 1 · 2016-12-16T01:03:56

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#