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About LinkBacksFInd the value of $\displaystyle \lim_{x\to\frac{\pi}{4}} \frac{tan\:2x}{cot(\frac{\pi}{4}-x)}$
Thanks!
Let $\displaystyle u=\frac{\pi}{4}-x$
$\displaystyle L=\lim_{x\rightarrow{\frac{\pi}{4}}}\frac{\tan(2x) }{\cot(\frac{\pi}{4}-x)}=\lim_{u\rightarrow{0}}\frac{\tan(2(\frac{\pi}{ 4}-u))}{\cot(u)}$
$\displaystyle \tan(2x)=\frac{2\tan(x)}{1-\tan^2(x)}$
$\displaystyle L=\lim_{u\rightarrow{0}}\frac{\tan(2(\frac{\pi}{4}-u))}{\cot(u)}=\lim_{u\rightarrow{0}}\frac{\frac{2\ tan(\frac{\pi}{4}-u)}{1-\tan^2(\frac{\pi}{4}-u)}}{\cot(u)}$
$\displaystyle L=\lim_{u\rightarrow{0}}\frac{\frac{2\tan(\frac{\p i}{4}-u)}{1-\tan^2(\frac{\pi}{4}-u)}}{\cot(u)}=\lim_{u\rightarrow{0}}\frac{\frac{2\ tan(\frac{\pi}{4}-u)}{1-\tan^2(\frac{\pi}{4}-u)}}{\frac{\cos(u)}{\sin(u)}}=2\lim_{u\rightarrow{ 0}}\frac{\sin(u)}{1-\tan^2(\frac{\pi}{4}-u)}$
$\displaystyle L=2\lim_{u\rightarrow{0}}\frac{\sin(u)}{1-\tan^2(\frac{\pi}{4}-u)}=2\lim_{u\rightarrow{0}}\frac{\sin(u)}{(1-\tan(\frac{\pi}{4}-u))(1+\tan(\frac{\pi}{4}-u))}$
$\displaystyle L=\lim_{u\rightarrow{0}}\frac{\sin(u)}{1-\tan(\frac{\pi}{4}-u)}=\frac{\sqrt[]{2}}{2}\lim_{u\rightarrow{0}}\frac{\sin(u)}{\cos(\ frac{\pi}{4}-u)-\sin(\frac{\pi}{4}-u)}$
Now recall the difference formulas: $\displaystyle \cos(\frac{\pi}{4}-u)=\frac{\sqrt[]{2}}{2}(\cos(u)+\sin(u))$
And: $\displaystyle \sin(\frac{\pi}{4}-u)=\frac{\sqrt[]{2}}{2}(\cos(u)-\sin(u))$
Thus: $\displaystyle L=\frac{\sqrt[]{2}}{2}\lim_{u\rightarrow{0}}\frac{\sin(u)}{\frac{ \sqrt[]{2}}{2}(2\sin(u))}\lim_{u\rightarrow{0}}\frac{\sin (u)}{2\sin(u)}=\frac{1}{2}$
Do you know l'Hôpital's rule? If your limit calculations give you $\displaystyle \frac {0}{0}$ or $\displaystyle \frac {\infty}{\infty}$, you can differentiate the numerator and denominator to get a result. In this question, you have to differentiate them a few times, just keep doing it until you get a result except $\displaystyle \frac {0}{0}$ or $\displaystyle \frac {\infty}{\infty}$.
Do you know that L'Hôpital's Rule does not belong to Guillaume de L'Hôpital?Originally Posted by wingless
Do you know that it's more interesting solvin' limits without such Rule?
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L'Hôpital's Rule is boring! I'd use it when extremely be necessary. (For a really, really, really hard limit.)
Do you know that L'Hopital's rule is for little girls who are afraid to calculate limits? Because when you solve without this rule you got to come up with bound and approximates for the function, which is what a large portion of analysis is about. Thus, solving without L'Hopital teaches you these things.
Just write,
$\displaystyle \lim_{u\to 0}\frac{\tan 2\left( \frac{\pi}{4} - u\right)}{\cot u} = \lim_{u\to 0}\frac{\tan \left( \frac{\pi}{2} - 2u \right)}{\cot u} = \lim_{u\to 0}\frac{\cot 2u}{\cot u} = \lim_{u\to 0}\frac{\tan u}{\tan 2u} = \frac{1}{2}$
Because,
$\displaystyle \frac{\tan u}{\tan 2u} = \frac{\sin u}{\sin 2u} \cdot \frac{\cos 2u}{\cos u}$
I’m just a student...I’m not a mathematician like you guys...I just care about knowing how to solve a question(while learning why) but if I can use l'Hopital's rule and it takes me 1 min. vs 10 min. doing it the "pro" way...I’ll take the little girls' way. As long as I can solve the problem and get the marks...I’m good.
...i did do that but then i got stuck when i got to $\displaystyle \frac{tan (\frac{\pi}{2}-2u)}{cot\:u}$...i didnt know you could just simplfy to $\displaystyle cot\:2u$
...I knew you could simplfy $\displaystyle tan(\frac{\pi}{2}-x)$ to $\displaystyle cot\:x$...i guess i dont know my trig well enough
That means you have a lot of time to solve every limit without the rule, I don't. Most of the students don't have much time either. I know, this rule 'simplifies' the calculations and makes them less significant. But please don't say "L'Hopital is for noobs", "only little girls use it" etc. People can choose their way to go, if they know the advantages and disadvantages of them.
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