Solving a Limit Involving tanx and pi/4

  • Thread starter Thread starter Glissando
  • Start date Start date
  • Tags Tags
    Limit
Glissando
Messages
34
Reaction score
0

Homework Statement


Find the limit:

lim (tanx-1)/(x-pi/4)
x->pi/4


Homework Equations


tanx = sinx/cosx


The Attempt at a Solution



lim (sinx/cosx-1)/(x-pi/4)
x->pi/4

lim [(sinx-cosx)/cosx]/(x-pi/4)
x->pi/4

I have no idea what to do after this ): I also tried squaring the whole function and getting tan2x-1 = sec2x, but I get so lost on what to do with pi/4 because it just keeps becoming undefined!

Thanks for your help!
 
Physics news on Phys.org
Hi Glissando! :smile:

I take it you're not allowed to use L'hopitals rule?? In that case, I would first write

\tan(x)=\tan((x-\pi/4)+\pi/4)

and work that out. That way you can write everything in function of x-\pi/4.
 
micromass said:
Hi Glissando! :smile:

I take it you're not allowed to use L'hopitals rule?? In that case, I would first write

\tan(x)=\tan((x-\pi/4)+\pi/4)

and work that out. That way you can write everything in function of x-\pi/4.

Hi micromass,

Thanks for your quick response! I'm not too sure what you mean by working it out...do I plug that back into the original equation? Am I solving for x?

Thanks!
 
Well, you know formula's for \tan(\alpha+\beta). So apply these formula's on

\tan((x-\pi/4)+\pi/4)

Our goal is to write everything in function of x-\pi/4
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top