lim x->0 1-cosx/x
To find the limit of the function as x approaches 0, we can apply the special limit known as the limit as x approaches 0 of (1-cos(x))/x
To find the limit of the function as x approaches 0, we can apply the special limit known as the limit as x approaches 0 of (1-cos(x))/x.
Let’s start by simplifying the function:
lim x->0 (1 – cos(x))/x
To simplify further, we’ll multiply the numerator and denominator by 1 + cos(x):
lim x->0 (1 – cos(x))/x * (1 + cos(x))/(1 + cos(x))
Expanding this expression, we get:
lim x->0 (1 – cos(x))(1 + cos(x))/(x(1 + cos(x)))
Now, using the difference of squares formula (a^2 – b^2 = (a + b)(a – b)), we can simplify the numerator:
lim x->0 (1^2 – cos^2(x))/(x(1 + cos(x)))
This simplifies to:
lim x->0 (sin^2(x))/(x(1 + cos(x)))
Next, we can write sin^2(x) as (sin(x))^2:
lim x->0 (sin(x))^2/(x(1 + cos(x)))
Now, we can use a trigonometric identity, sin^2(x) = (1 – cos(2x))/2, to further simplify the expression:
lim x->0 [(1 – cos(2x))/2]/(x(1 + cos(x)))
Simplifying the expression, we get:
lim x->0 (1 – cos(2x))/(2x(1 + cos(x)))
Notice that now the limit is in indeterminate form 0/0. To evaluate this limit, we can use L’Hôpital’s rule, which states that if we have a limit in the form 0/0 or ∞/∞, we can take the derivative of the numerator and the derivative of the denominator separately, and then evaluate the limit again.
Taking the derivative of the numerator and denominator, we get:
= lim x->0 (-2sin(2x))/(2(1 + cos(x)) – 2x(-sin(x)))
= lim x->0 -sin(2x)/(1 + cos(x) – x(sin(x)))
Now, applying the limit once again, we have:
= -sin(2(0))/(1 + cos(0) – 0(sin(0)))
= -sin(0)/(1 + 1 – 0)
= 0/(2)
= 0
Therefore, the limit as x approaches 0 of (1 – cos(x))/x is equal to 0.
More Answers:
[next_post_link]