How do you evaluate #lim_(x-> 0) tanx/x#?

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Answer 1 · 2017-08-24T23:50:11

Since #tan(0) = sin(0)/cos(0) =0/1 = 0# and the denomianator equals #0#, we can use L'Hospitals rule .

#L = lim_(x->0) (sec^2x)/1#

#L = lim_(x->0) sec^2x#

#L = sec^2(0)#

#L = 1/cos^2(0)#

#L = 1#

If we check the graph of the function #f(x) = tanx/x#, we see that when #x -> 0#, #y# approaches #1#.

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Hopefully this helps!

Answer 2 · 2017-08-25T01:55:30

Please see below.

#tanx = sinx/cosx# so

#tanx/x = sinx/(xcosx) = sinx/x * 1/cosx#.

So,

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

# = lim_(xrarro) sinx/x * lim_(xrarro)1/cosx#

# = 1*1/1 = 1#