How do you find the limit #(cosx-1)/sinx# as #x->0#?

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Answer 1 · 2016-11-12T15:13:18

Use the two fundamental trigonometric limts (and algebra).

#lim_(xrarr0)sinx/x = lim_(xrarr0) x/sinx = 1#

#lim_(xrarr0)(cosx-1)/x = lim_(xrarr0)(1-cosx)/x = 0#

Note that

#(cosx-1)/sinx = (cosx-1)/sinx * x/x#

# = (cosx-1)/x * x/sinx#

Since the limits of both factors exist, the limit of the product is the product of the limits. So

#lim_(xrarr0) (cosx-1)/sinx = lim_(xrarr0)(cosx-1)/x * lim_(xrarr0) x/sinx#

# = (0)*(1) = 0#