find the value...
lim (tan 2x- x)/(3x- sin x)
x>0
To find the limit of the given expression, lim (tan 2x - x)/(3x - sin x) as x approaches 0 from the positive side, we can use L'Hôpital's rule.
L'Hôpital's rule states that if we have a limit of the form 0/0 or ∞/∞, and the limit of the derivative of the numerator over the derivative of the denominator exists (either as a finite number or ∞), then the limit of the original expression is equal to the limit of the derivative of the numerator over the derivative of the denominator.
Let's find the derivatives of the numerator and the denominator:
f'(x) = d/dx (tan 2x - x) = sec^2(2x) - 1
g'(x) = d/dx (3x - sin x) = 3 - cos x
Now, we can rewrite the original limit using L'Hôpital's rule:
lim (tan 2x - x)/(3x - sin x) = lim (sec^2(2x) - 1)/(3 - cos x)
Now, we can evaluate the limit by plugging in x = 0:
lim (sec^2(2x) - 1)/(3 - cos x) = lim (sec^2(2 * 0) - 1)/(3 - cos 0)
= lim (sec^2(0) - 1)/(3 - 1)
= lim (1 - 1)/(3 - 1)
= lim 0/2
= 0
Therefore, the value of the given limit as x approaches 0 from the positive side is 0.
plz answer without L hospital rule
using the well-known proofs that sinx -> x, we can conclude that
tan2x -> 2x
so we have the limit as
(2x-x)/(3x-x) = 1/2