3Solve Find the Limit Homework Statement

  • Thread starter Thread starter brycenrg
  • Start date Start date
  • Tags Tags
    Limit
brycenrg
Messages
95
Reaction score
2

Homework Statement


Find the limit
lim (sin(3x))/x
x-->0


Homework Equations


There is two steps which I don't understand how they work, or why they work. Ill show you.
I'm using my answer book to help me out, but it doesn't give a satisfactory explanation.


The Attempt at a Solution


lim (sin(3x))/x
x-->0

= lim 3(sin(3x))/3x
x-->0

= 3lim (sin(3x))/3x as x -> 0 3x -> 0 Why do I do this maneuver? I understand its for
x-->0 substitution

= 3lim (sin(3x))/3x let theta = 3x

= 3lim (sin(theta))/theta

= 3(1) Why does lim (sin(theta))/theta equal 1?
 
Physics news on Phys.org
brycenrg said:
Why does lim (sin(theta))/theta equal 1?

Here's a geometrical argument that 'proves' the limit is equal to 1: http://math.ucsd.edu/~wgarner/math20a/sin%28x%29_over_x.htm .

I think this argument is enough for your proof, you're probably not supposed to deduce the limit from the Maclaurin series expansion of the sine function (which would be the rigorous way to do it).
 
Last edited by a moderator:
There are many ways to check it... first of all you have a form of 0/0 for which you can use the del'hopitale...
Another way is the Taylor expansion of sine : the argument is small so you just need the first order term which is \theta and gives 1 with the denominator...
Another way is via the sandwich rule demonstrated by hilbert12's reference.
 
Thank you very much. I like to learn how it works for some odd reason, ocd or something. Its like i doubt maths even though its been around for ever lol. Thanks.
 
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...
Back
Top