Question

Question #48bca

Answer 1

`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`