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 3,255 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) limh→0cos(c+h)=\displaystyle \lim_{h \to 0} cos(c + h) =h→0limcos(c+h)= limh→0[cos(c)cos(h)−sin(c)sin(h)]=\displaystyle \lim_{h \to 0}[cos(c)cos(h) - sin(c)sin(h)] =h→0lim[cos(c)cos(h)−sin(c)sin(h)]=. . . .I left off here.\displaystyle . \ . \ . \ . I \ left \ off \ here.. . . .I left off here. Ask yourself as to what limh→0cos(h)\displaystyle \lim_{h \to 0}cos(h)h→0limcos(h) is and what limh→0sin(h)\displaystyle \lim_{h \to 0}sin(h)h→0limsin(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) limh→0cos(c+h)=\displaystyle \lim_{h \to 0} cos(c + h) =h→0limcos(c+h)= limh→0[cos(c)cos(h)−sin(c)sin(h)]=\displaystyle \lim_{h \to 0}[cos(c)cos(h) - sin(c)sin(h)] =h→0lim[cos(c)cos(h)−sin(c)sin(h)]=. . . .I left off here.\displaystyle . \ . \ . \ . I \ left \ off \ here.. . . .I left off here. Ask yourself as to what limh→0cos(h)\displaystyle \lim_{h \to 0}cos(h)h→0limcos(h) is and what limh→0sin(h)\displaystyle \lim_{h \to 0}sin(h)h→0limsin(h) is. Then continue where I left off.
D Deleted member 4993 Guest Feb 17, 2011 #3 Another way: From definition cosh(ϕ) = eϕ + e−ϕ2\displaystyle cosh(\phi) \ = \ \frac{e^{\phi} \ + \ e^{-\phi}}{2}cosh(ϕ) = 2eϕ + e−ϕ Now express cosh(?+h) and take the limit
Another way: From definition cosh(ϕ) = eϕ + e−ϕ2\displaystyle cosh(\phi) \ = \ \frac{e^{\phi} \ + \ e^{-\phi}}{2}cosh(ϕ) = 2eϕ + e−ϕ Now express cosh(?+h) and take the limit
