Question

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

Answer 1

-10

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

Answer 2

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

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-(x^3)/(3!)\right)\\\\\text{So, in the neighbourhood of 0}\\\\\begin{aligned}(sin(4x)-4sin(x)) &\sim \left( 4x-((4x)^3)/(3!)-4x+(4x^3)/(3!)\right)\\\\&\sim \left((x^3*4*(1-4^2))/(3*2)\right)\\\\&\sim \left((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) \ {(sin(4x)-4sin(x))/(x^3)}\\\\= \lim_(x\rightarrow0) \ {(-10*x^3)/(x^3)}\\\\=-10`

Thank you