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

2 Answers
Aug 24, 2017

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!

Aug 25, 2017

Please see below.

Explanation:

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