# Lim (sinx-cosx)/(pi-4x) where x->pi/4

1. Jan 6, 2010

### danculax

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

Hi! Could someone help me to solve this problem without using L'Hospital rule? Any hint would be splendid...

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

2. Jan 6, 2010

### ideasrule

Re: lim (sinx-cosx)/(pi-4x) where x-->pi/4

L'Hopital's rule?

3. Jan 6, 2010

### danculax

Re: lim (sinx-cosx)/(pi-4x) where x-->pi/4

Sorry, but without L'Hospital. I forgot to mention...With L'Hospital is easy to solve and it gives
=-sqrt(2)/4

4. Jan 6, 2010

### rock.freak667

Re: lim (sinx-cosx)/(pi-4x) where x-->pi/4

Try rewriting sin(x)-cos(x) in the form Rsin(x-A)

5. Jan 7, 2010

### Bohrok

Re: lim (sinx-cosx)/(pi-4x) where x-->pi/4

You could multiply by (cosx + sinx)/(cosx + sinx) and use the identity cos2x -sin2x = cos2x. Then let u = 2x, use the identity sin($\pi$/2 - x) = cosx, rewrite it some more, and use the common trig limit sinx/x. Don't forget to change the limits with each substitution.

6. Jan 7, 2010

### danculax

Re: lim (sinx-cosx)/(pi-4x) where x-->pi/4

Wow! Thank you for this great tip! I tried a similar method but I had problems to find a good substitution. u=2x really works and the problem is solved!!! Thanks again!

7. Jan 7, 2010

### danculax

Re: lim (sinx-cosx)/(pi-4x) where x-->pi/4

Now I have two solutions for my limit problem. I figured out how to transform sin(x)-cos(x) into Rsin(x-A) form:

R=Sqrt(2)
A=pi/4

sinx-cosx=Sqrt(2)*sin(x-pi/4)

8. Jan 9, 2010

### danielatha4

Re: lim (sinx-cosx)/(pi-4x) where x-->pi/4

L'Hopital's rule is simple. It says that if the limit of a quotient takes on an indeterminate form, such as 0/0 then the limit is equivalent to the derivative of the numerator divided by the derivative of the denominator.

sinx-cosx / pi -4x

set the same limit to

cosx+sinx / -4

9. Jan 9, 2010

### TheoMcCloskey

Re: lim (sinx-cosx)/(pi-4x) where x-->pi/4

Alternatively, you could expand $\sin(x)$ and $\sin(x)$ in a Taylor's Series about $x = \pi/4$. The numerator subtraction will eliminate even power terms and the denomiator will cancel one power of $(x-\pi/4)$ term. The result will have a leading constant term which is your limit as $x\rightarrow \pi/4$.

10. Jan 9, 2010

### danielatha4

Re: lim (sinx-cosx)/(pi-4x) where x-->pi/4

Hmm, I'm not familiar with expanding in a Taylor's Series.