find the lim (cos x-1)/ (sin 2x)
x->0
can you use L'Hopital's rule?
if so
lim (cos x-1)/ (sin 2x)
x->0
= lim (-sinx)/(2cos 2x) as x ---> 0
= 0/2 = 0
no u cant
some common trig limits you should have come across:
lim( (sinx)/x ) as x --> 0 = 1
lim( x/sinx ) as x --> 0 = 1
lim( (cosx - 1)/x) as x -- 0 = 0
so lets rewrite the question as
lim[ ((cos x-1)/x) * x(2sinxcosx) ]
= lim( (cosx - 1)/x) * lim (x/sinx) * lim (1/2cosx)
= 0(1)(1/2) = 0
To find the limit of the given expression as x approaches 0, we can simplify the expression and then substitute the value of x.
Step 1: Simplify the expression.
We can use the trigonometric identity sin(2x) = 2sin(x)cos(x) to simplify the expression.
(cos(x) - 1) / (sin(2x)) = (cos(x) - 1) / (2sin(x)cos(x))
Step 2: Factor out common terms.
(cos(x) - 1) / (2sin(x)cos(x)) = (cos(x) - 1) / (2sin(x) * cos(x)) = (cos(x) - 1) / (2sin(x)) * (1/cos(x))
Step 3: Remove the common factor.
(1/cos(x)) is the reciprocal of cos(x), which goes to infinity when cos(x) goes to 0, so we can remove the common factor from the equation.
(cos(x) - 1) / (2sin(x)) * (1/cos(x)) = (cos(x) - 1) / (2sin(x))
Step 4: Substitute the value of x.
Now, we can substitute the value of x=0 into the expression.
(cos(0) - 1) / (2sin(0)) = (1 - 1) / (0) = 0/0
Step 5: Evaluate the limit.
The expression simplifies to 0/0, which is an indeterminate form. This means that we cannot directly evaluate the limit using substitution. We need to apply a different method, such as L'Hopital's Rule or other algebraic manipulations.
Applying L'Hopital's Rule, we take the derivative of the numerator and denominator separately and then evaluate the limit again.
Differentiating the numerator: d/dx [cos(x) - 1] = -sin(x)
Differentiating the denominator: d/dx [2sin(x)] = 2cos(x)
Now we have:
(-sin(x)) / (2cos(x))
Substituting x=0 into this expression:
(-sin(0)) / (2cos(0)) = (0) / (2) = 0
Therefore, the limit of (cos(x) - 1) / (sin(2x)) as x approaches 0 is 0.