Can some one please explain this problem to me?
Find lim x->0 (sin5x/sin7x)
recall that
lim sinØ/Ø = 1 , as Ø----> 0
so we want to make our expression "look like that"
suppose we multiply sin 5x/sin 7x by (7x/5x)*(5x/7x)
so we get
limit [ (sin 5x/(5x) / sin 7x/(7x) ]*(5x)/(7x)
as x---> 0
of course as x---> 0 , so does 5x as well as 7x
= 1/1 * 5/7
= 5/7
you can test this with your calculator,
set it to radians, (the above is true in radians, NOT in degrees)
let x = something like .000001 and evaluate
sin 5x/sin 7x
you will get .714285714
and 5/7 = .714285714 , how about that!
To find the limit of (sin5x/sin7x) as x approaches 0, you can use the limit properties and the fact that sin(x)/x approaches 1 as x approaches 0.
Step 1: Rewrite the expression using the sine of a double angle identity:
lim x->0 (sin5x/sin7x) = lim x->0 [(sin(2x + 3x))/sin7x]
Step 2: Use the double angle formula sin(2x) = 2sin(x)cos(x):
= lim x->0 [(2sin(x)cos(x) + sin(3x))/sin7x]
Step 3: Split the limit into two separate limits using the sum/difference rule:
= lim x->0 [2sin(x)cos(x)/sin7x] + lim x->0 [sin(3x)/sin7x]
Step 4: Apply the limit property sin(x)/x approaches 1 as x approaches 0:
= 2 * 1/7 + 0
Step 5: Simplify the expression:
= 2/7
So, the limit of (sin5x/sin7x) as x approaches 0 is 2/7.
To find the limit of the function (sin5x/sin7x) as x approaches 0, you can use the basic limit properties and trigonometric identities.
Let's start by applying the trigonometric identity: sin(A)/sin(B) = (2sin((A-B)/2)cos((A+B)/2)).
Using this identity, we can rewrite the function as:
(sin5x/sin7x) = (2sin((5x-7x)/2)cos((5x+7x)/2)) = (2sin(-x/2)cos(6x/2)) = (-2sin(x/2)cos(3x))
Now, as x approaches 0, both sin(x/2) and cos(3x) approach 0. So, we have (-2 * 0 * 0), which is equal to 0.
Hence, the limit of (sin5x/sin7x) as x approaches 0 is 0.