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

[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

Thanks in advance!

 

[ideasrule]

L'Hopital's rule?

 

[danculax]

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

 

[rock.freak667]

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

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

 

[Bohrok]

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.

 

[danculax]

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.

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!

 

[danculax]

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

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)

Thank you for your help!

 

[danielatha4]

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

 

[TheoMcCloskey]

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 $x\rightarrow \pi/4$.

 

[danielatha4]

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