I have a definition lim theta approaching 0 sin theta/theta = 1
so show if the limit x approaching 0 sin2x/5=2/5
well the point is that sin x ---> x -x^2/2 + .......
as x------> 0
sin x ---> x
so
sin(2x) / 5x ----> [2x - 4x^2 ...]/5x
as x---> 0
sin(2x) /5 ---> 2x/5x = 2/5
you know that
lim(u->0) sinu/u = 1
let u = 2/5 x
lim sin2x/5x
= lim sinu/(5/2 u)
= lim sinu/u * 2/5
= 1 * 2/5
= 2/5
To evaluate the limit of sin(2x)/5 as x approaches 0, we can use the double-angle identity for sine and the limit property that the limit of a constant multiple of a function equals the constant multiplied by the limit of the function. Here are the steps to solve it:
Step 1: Apply the double-angle identity for sine: sin(2x) = 2sin(x)cos(x).
Step 2: Rewrite the expression: sin(2x)/5 = (2sin(x)cos(x))/5.
Step 3: Separate the two terms in the numerator: (2/5)(sin(x)cos(x)).
Step 4: Recognize that sin(x)/x has a limit of 1 as x approaches 0 based on the given definition.
Step 5: Apply the limit property that states, if lim(x→a) f(x) = L and lim(x→a) g(x) = M, then lim(x→a) (f(x)g(x)) = L*M.
Step 6: Substitute the limit of sin(x)/x as x approaches 0 and simplify: (2/5)(1*cos(0)) = 2/5.
Therefore, the limit of sin(2x)/5 as x approaches 0 is 2/5.