How do you evaluate the limit #sin(5x)/x# as x approaches #0#?

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Answer 1 · 2016-11-07T21:45:12

Use #lim_(thetararr0)sintheta/theta = 1# and some other tools.

#lim_(xrarr0)sin(5x)/x#

We'd like to use the limit mentioned above, so we need to have #theta = 5x#

To get #5x# in the denominator, we'll multiply by #5/5#

#lim_(xrarr0)sin(5x)/x = lim_(xrarr0)(5sin(5x))/(5x) #

Now factor the #5# in the numerator outside the limit.

# = 5lim_(xrarr0)sin(5x)/(5x) #

As #xrarr0#, #5xrarr0# so we have:

# = 5(1) = 5#

(The limit is #5#.)