How do you find the limit of #sec (x-1) / (x sec x) # as x approaches 0?

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Answer 1 · 2017-08-08T14:45:45

The limit doesn't exist.

Call the limit #L#. We can rewrite as follows.

#L = cosx/(xcos(x - 1))#

Because #secx = 1/cosx#. If we evaluate, we get

#L = 1/0 = oo#

But, if we check to see if the left and right-hand limits are equal we get

#lim_(x-> 0^-) = -oo and lim_(x->0^+) = +oo#

Since the left-hand and right-hand limits aren't equal, we can conclude that this limit doesn't exist.

Hopefully this helps!