## MarcLeclair Group Title Use L'Hostpial rule to find the following limit: Lim x--> 0 of 6x/7arctan(4x) I did the work And i get the following 6(1+96x^2) / 7 and i end up with 6/7 however that's not the right answer, any insight on where I went wrong? one year ago one year ago

1. Yahoo! Group Title

Differentiate Numerator and Denominator Seperately...)

2. MarcLeclair Group Title

This is what I did..... top gives 6 and denominator gives 7 1/1-(4x)^2... therefore you get 6 ( 1-(4x)^2)/7 no?

3. ZeHanz Group Title

l'Hopital's rule:$\lim_{x \rightarrow0}\frac{ 6x }{ 7\arctan4x }=$$\lim_{x \rightarrow 0}\frac{ (6x)' }{ (7arctan4x)' }=$$\lim_{x \rightarrow 0}\frac{ 6 }{ 7* \frac{ 1 }{ 1+(4x)^2}*4 }=$$\frac{ 6 }{ 7*1*4 }=\frac{ 6 }{ 28 }=\frac{ 3 }{ 14 }$ You have to multiply with 4 in the denominator, because of the Chain Rule... Hope this helps! ZeHanz

4. MarcLeclair Group Title

I forgot the chain rule... god thank you so much! that was really helpful, remembering that trig function are included in the chain rule just made me solve half my problems again! have a good day

5. ZeHanz Group Title

You are welcome!