Question

What is the limit of `(4x − sin4x)/(4x − tan4x)` as x approaches 0?

Answer 1

1

Explanation:

Calling `a(x) = 4x`, `b(x) = -sin(4x)` and `c(x)=cos(4x)` we
have `lim_{x->infty}(4x-sin(4x))/(4x-tan(4x))equiv lim_{x->0}((c(x)(a(x)+b(x)))/(c(x)a(x)+b(x)))` but
`lim_{x->x_0}((c(x)(a(x)+b(x)))/(c(x)a(x)+b(x))) =(c(x_0)(a(x_0)+b(x_0)))/(c(x_0)a(x_0)+b(x_0))` But in this case `x_0=0` and `c(x) = cos(4x)` so `c(x_0)=1`. Putting all together we have
`lim_{x->x_0}((c(x)(a(x)+b(x)))/(c(x)a(x)+b(x))) = (a(x_0)+b(x_0))/(a(x_0)+b(x_0))=1`

Answer 2

`lim_(xrarr0)(4x-4sinx)/(4x-tan4x) = -1/2`

Explanation:

`lim_(xrarr0)(4x-sin4x)/(4x-tan4x)` has indeterminate initial form `0/0`.

I'll use l'Hospital's rule.

`lim_(xrarr0)(4x-sin4x)/(4x-tan4x) = lim_(xrarr0)(4-4cos4x)/(4-4sec^2 4x)`

`= lim_(xrarr0)(1-cos4x)/(1-sec^2 4x)`

The form is still `0/0`, so we'll use L'Hospital again. Note that to differentiate `sec^2 4x` we'll use the chain rule twice.

`= lim_(xrarr0)(4sin4x)/(-2sec(4x)[sec(4x)tan(4x)(4)])`

`= lim_(xrarr0)(4sin4x)/(-8sec^2(4x)tan(4x))`

`= lim_(xrarr0)(-sin4x)/(2sec^2(4x)(sin(4x))/cos(4x))`

`= lim_(xrarr0)(-1)/(2sec^3(4x))`

`= lim_(xrarr0)(-cos^3 (4x))/2` (if you prefer)

`= -1/2`

Here is the graph:

graph{(4x-sin(4x))/(4x-tan(4x) [-4.375, 3.42, -1.795, 2.1]}