Question #48bca

1 Answer
Anthony R.
Jan 12, 2018

#lim_(x->0)(tan3x)/(2x)=3/2#

Explanation:

Did You mean #lim_(x->0)(tan3x)/(2x)?#

If so, then...

Let's rewrite the limit (though not necessary) as

#1/2*lim_(x->0)(tan3x)/(x)#

If we use direct substitution we will end up with an indeterminate form #0/0# because:

The #lim_(x->0)tan(x)=0# can easily be proven by looking at its graph:

graph{tanx [-10, 10, -5, 5]}

And plugging in #0# for #x# is simply #0#

#:.lim_(x->0)(tan(3x))/(2x)=0/0#

Given that, it should become clear that we can apply L'Hospital's Rule which means we take the derivative of both the numerator and denominator separately and then attempt to substitute #0# again and evaluate the limit.

So,

#1/2*lim_(x->0)=1/2*lim_(x->0)(d/dx[tan(3x)])/(d/dx[x])=1/2*lim_(x->0)(3sec^2(3x))/1#

#=1/2*lim_(x->0)3sec^2(3x)=1/2*3/cos^2(3*0)=1/2*3/1=3/2#

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