What is the limit of (sin2s)^4/s^4 as s approaches 0?

  • Thread starter Thread starter beneakin
  • Start date Start date
  • Tags Tags
    Limit Trig
beneakin
Messages
13
Reaction score
0

Homework Statement


Find the limit or prove it does not exist.

lim as s->0 (sin2s)^4/s^4


Homework Equations


i know i have to use sinx/x = 1 but I am having trouble manipulating the function into that form


The Attempt at a Solution



my first thought was that i could just take the 2 out as a constant would that just be (2^4)(sinx)^4/x^4

but then i have (sinx)^4 not sinX^4 I am stuck on what king of algebra will work here
 
Physics news on Phys.org
a^2/b^2=(a/b)^2
 
Lim_(x->0) of sin(ax)/(ax)=1
 
ahh so then (2*1)^4 = 16

thanks alot
 
beneakin said:
ahh so then (2*1)^4 = 16

thanks alot
good, welcome!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

Similar threads

Back
Top