lim h->0 of cos (c+h) = cos(c) how do you prove this ?
K kelsjo813 New member Joined Feb 16, 2011 Messages 1 Feb 16, 2011 #1 lim h->0 of cos (c+h) = cos(c) how do you prove this ?
L lookagain Elite Member Joined Aug 22, 2010 Messages 3,230 Feb 17, 2011 #2 kelsjo813 said: lim h->0 of cos (c+h) = cos(c) how do you prove this ? Click to expand... From a page of trig formulas, for angles A and B, cos(A + B) = cos(A)cos(B) - sin(A)sin(B) \(\displaystyle \lim_{h \to 0} cos(c + h) =\) \(\displaystyle \lim_{h \to 0}[cos(c)cos(h) - sin(c)sin(h)] =\)\(\displaystyle . \ . \ . \ . I \ left \ off \ here.\) Ask yourself as to what \(\displaystyle \lim_{h \to 0}cos(h)\) is and what \(\displaystyle \lim_{h \to 0}sin(h)\) is. Then continue where I left off.
kelsjo813 said: lim h->0 of cos (c+h) = cos(c) how do you prove this ? Click to expand... From a page of trig formulas, for angles A and B, cos(A + B) = cos(A)cos(B) - sin(A)sin(B) \(\displaystyle \lim_{h \to 0} cos(c + h) =\) \(\displaystyle \lim_{h \to 0}[cos(c)cos(h) - sin(c)sin(h)] =\)\(\displaystyle . \ . \ . \ . I \ left \ off \ here.\) Ask yourself as to what \(\displaystyle \lim_{h \to 0}cos(h)\) is and what \(\displaystyle \lim_{h \to 0}sin(h)\) is. Then continue where I left off.
D Deleted member 4993 Guest Feb 17, 2011 #3 Another way: From definition \(\displaystyle cosh(\phi) \ = \ \frac{e^{\phi} \ + \ e^{-\phi}}{2}\) Now express cosh(?+h) and take the limit
Another way: From definition \(\displaystyle cosh(\phi) \ = \ \frac{e^{\phi} \ + \ e^{-\phi}}{2}\) Now express cosh(?+h) and take the limit
