Question

Limit (sin4x-4sinx)/x^3 when x close to zero

Answer 1

$\Large \boxed{\sf \bf \ \ \lim_{x\rightarrow0} \ {\dfrac{sin(4x)-4sin(x)}{x^3}}=-10 \ \ }$

Step-by-step explanation:

Hello, please consider the following.

Using Maclaurin series expansion, we can find an equivalent of sin(x) in the neighbourhood of 0.

$sin(x) \sim \left(x-\dfrac{x^3}{3!}\right)\\\\\text{So, in the neighbourhood of 0}\\\\\begin{aligned}(sin(4x)-4sin(x)) &\sim \left( 4x-\dfrac{(4x)^3}{3!}-4x+\dfrac{4x^3}{3!}\right)\\\\&\sim \left(\dfrac{x^3*4*(1-4^2)}{3*2}\right)\\\\&\sim \left(\dfrac{x^3*2*(-15)}{3}\right)\\\\&\sim \left(x^3*2*(-5)\right)\\\\&\sim \left(x^3*(-10)\right)\\\end{aligned}$

Then,

$\displaystyle \lim_{x\rightarrow0} \ {\dfrac{sin(4x)-4sin(x)}{x^3}}\\\\= \lim_{x\rightarrow0} \ {\dfrac{-10*x^3}{x^3}}\\\\=-10$

Thank you

Answer 2

Answer:

-10

Step-by-step explanation:

L' Hopital's rule can be used to solve it.

We search for lim (f(x)/g(x))

this is the same as lim (f' (x)/g' (x))

this is the same as lim f'' (x)/g'' (x)

so here we have f(x) =sin(4x)-4sin(x)

g(x)=x^3

g'' (x)= 6x

f'' (x)=-16sin(4x)+4sin(x)

so lim f'' (x) / g ''(x) = -16/6 lim sin(4x)/x + 4/6 lim (sin(x)/x)

= -16*4/6+4/6

=(-64+4)/6=-60/6=-10