How do you evaluate the limit #sinx/tanx# as x approaches #0#?

Question

1798 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-21T12:46:01

#Lt_(x->0)sinx/tanx=1#

#Lt_(x->0)sinx/tanx#

= #Lt_(x->0)sinx/(sinx/cosx)#

= #Lt_(x->0)sinx xxcosx/sinx#

= #Lt_(x->0)cosx#

= #cos0#

= #1#

Answer 2 · 2016-09-21T12:54:04

# lim_(xrarr0) sinx/tanx=1#.

We use the Standard Form of Limit :

#lim_(thetararr0)sintheta/theta=1, theta in RR#.

From the above, we have,

#lim_(thetararr0) tantheta/theta#

#=lim_(thetararr0) {sintheta/theta*1/costheta}#

#={lim_(thetararr0) sintheta/theta}*{lim_(thetararr0)1/costheta}#

#=1*(1/cos0)#

# :. lim_(thetararr0) tantheta/theta=1.#

Now, # lim_(xrarr0) sinx/tanx#

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

#={lim_(xrarr0) sinx/x}{lim_(xrarr0) x/tanx}#

#=1*1#

# :. lim_(xrarr0) sinx/tanx=1#.