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Limit of Trigonometric f'n as x approaches 0

  • Thread starter joshmdmd
  • Start date
  • #1
53
0

Homework Statement



Find the following limit as x tends to zero
(idk how to code this stuff properly so the lim is lim as x->0)
lim (sinx-tanx)/x^3


Homework Equations



lim(sinx)/x = 1
lim(cosx) = 1
lim(1-cosx)/x = 0

The Attempt at a Solution



attempted to multiply by conjugate
attempted to remove the root from the top and the bottom
no idea how to do this.
 

Answers and Replies

  • #2
307
3
I'm not sure what you are allowed to do here, but one way to solve this limit would be to consider the Taylor series of sin and tan about the point x = 0.

In particular, [itex]\sin(x) \approx x - x^3/3! + x^5/5! - ...[/itex] and [itex]\tan(x) \approx x + x^3/3 + 2x^5/5 + ...[/itex]. Hence, subtracting these two series, gives:

[tex]\sin(x) - \tan(x) \approx -(x^3/3! + x^3/3) + O(x^5)[/tex]

Substituting this into your function (i.e. dividing the above by [itex]x^3[/itex]) and taking the limit as that goes to zero will give you an answer.
 
  • #3
53
0
nope. it has to be done without L'Hospital's rule, and we learn the taylor / maclaurin series next semester?. My attempts to rearrange it only get down to x on the bottom, can't seem to shake it out.
 
  • #4
307
3
Okay... here is a slightly more cumbersome way,

Notice,
[tex]\frac{\sin(x) - \tan(x)}{x^3} = -\left(\frac{\sin(x)}{x}\right)^3\frac{1}{\cos^2(x) + \cos(x)}[/tex].

A simple application of limit laws will give you the solution from that stage.

Hint:
To obtain the above equality, start by multiplying the top and bottom of your fraction by [itex](\sin(x)+\tan(x))[/itex].
 

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