\lim_{x \to \frac{\pi}{4}} \frac{(\csc^2 x-2 \tan^2 x)}{(\cot x-1)} without L'Hôpitals

lugreget9

lugreget9

Answered question

2021-12-30

limxπ4(csc2x2tan2x)(cotx1) without L'Hôpitals Rule.

Answer & Explanation

Jenny Bolton

Jenny Bolton

Beginner2021-12-31Added 32 answers

Hint:
(csc2x2tan2x)(cotx1)=csc222tan2x+2cotx1
=csc2x2cotx1+2tanx1tan2x1tanx
=cot2x1cotx1+2tanx1tan2x1tanx
=cotx+1+2tanx(1+tanx)
It is easier to take the limit now
Serita Dewitt

Serita Dewitt

Beginner2022-01-01Added 41 answers

This limit can actually be calculated quite directly. You only need tanπ4=1
Rewrite the expression as follows and set t=tanx and consider t1:
(csc2x2tan2x)(cotx1)=cos2xsin2x+12tan2x1tanx1
=t=tanx1t2+12t21t1
=1+t22t4t(1t)
=(1t)(1+t)(1+2t2)t(1t)
=(1+t)(1+2t2)tt16
Vasquez

Vasquez

Expert2022-01-09Added 669 answers

I use (at the end)
cosxsinx=2((12)cosx(12)sinx)=2(sin(π4)cosxcos(π4)sinx)=2sin(π4x)v=limxπ4csc2x2tan2xcotx1=limxπ41sin2x2sin2xcos2xcosx

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