1use l hospital
1. \lim_{\theta \rightarrow \frac{\pi}{4}} \frac{\sqrt{2}-cos\theta -sin\theta}{(4\theta-\pi)^{2}}
2. \lim_{x\rightarrow 0}\frac{xtan2x-2xtanx}{(1-cos2x)^{2}}
3. \lim_{x\rightarrow 0}\frac{e^{x}-e^{xcosx}}{(x+sinx)}
I want to evaluate these without using expansions (if any). Please help
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9 Answers
71Arreyy. Pehla toh aise hi tha.. I could not do the rest with L' Hospital..
1(1)lim@→pi/4 √2-cos@-sin@(4@-pi)2
Applying De L Hospital's rule two times
lim@→pi/4cos@+sin@32
=116√2
1(2)limx→0 xtan2x - 2x tanx(1-cos2x)2
Put tan2x in terms of tanx,then convert it in the form of sin and cos.You will get
limx→02xcos2xsin4x
Applying De L hospital's rule
limx→0-4xsinxcosx+2cos2x4cos4x
=1/2.
1limx→0ex-excosx(x+sinx)
Applying Del Hospitals rule
limx→0ex-(-xsinx+cosx)excosx1+cosx
=0
Thank you!
1yup sorry... it is 1/16√2 for teh 1st one.... made a bit mistake,..
1708e^x-e^{xcosx}=e^{xcosx}(e^{x-xcosx}-1)\approx1.\lim_{x\to 0}\frac{(e^{x-xcosx}-1)}{x-xcosx}(x-xcosx)$\\\\ $\approx x(1-cosx)\approx x.(2sin^2\frac{x}{2})\approx x.\frac{2.x^2}{4}\approx \frac{x^3}{2}$\\\\ and $x+sinx\approx 2x$(at $x=0,sinx \approx x$)\\\\ So $\lim_{x\to 0}\frac{e^x-e^{xcosx}}{x+sinx}\approx \lim_{x\to 0}\frac{x^3}{2.2x}\approx 0$\\\\