nameVoid Messages 238 Reaction score 0 Thread starter Apr 8, 2009 #1 lim t-> 0 ,t^3/tan^3(2t) , not seeing nay identiites to solve with, escpeted to solve not using l hospitols
lim t-> 0 ,t^3/tan^3(2t) , not seeing nay identiites to solve with, escpeted to solve not using l hospitols
arildno Science Advisor Homework Helper Gold Member Dearly Missed Messages 10,165 Reaction score 138 Apr 8, 2009 #2 Well, you might try to rewrite this as: [tex]\lim_{t\to{0}}\frac{t^{3}}{\tan^{3}(2t)}=\lim_{t\to{0}}(\frac{t}{\sin(2t)})^{3}\cos^{3}(2t))=\lim_{t\to{0}}(\frac{1}{2})^{3}(\frac{2t}{\sin(2t)})^{3}\cos^{3}(2t))[/tex]
Well, you might try to rewrite this as: [tex]\lim_{t\to{0}}\frac{t^{3}}{\tan^{3}(2t)}=\lim_{t\to{0}}(\frac{t}{\sin(2t)})^{3}\cos^{3}(2t))=\lim_{t\to{0}}(\frac{1}{2})^{3}(\frac{2t}{\sin(2t)})^{3}\cos^{3}(2t))[/tex]
Gib Z Homework Helper Messages 3,341 Reaction score 7 Apr 8, 2009 #4 Really? What are the limits of those terms individually?
Gib Z Homework Helper Messages 3,341 Reaction score 7 Apr 8, 2009 #6 You've never learned [tex]\lim_{x\to 0} \frac{\sin x}{x} = 1[/tex] ?