Evaluate the limit below.limx→0 7x/sin(3x)
To evaluate the limit of the function (7x)/(sin(3x)) as x approaches 0, we can use L’Hôpital’s Rule
To evaluate the limit of the function (7x)/(sin(3x)) as x approaches 0, we can use L’Hôpital’s Rule. The rule states that if we have an indeterminate form of 0/0 or ∞/∞ when evaluating the limit of a function, we can take the derivative of the numerator and denominator separately and then evaluate the limit again.
Let’s apply L’Hôpital’s Rule to the given function. First, we need to find the derivatives of the numerator and denominator separately.
The derivative of 7x with respect to x is 7.
The derivative of sin(3x) with respect to x can be found using the chain rule. Let u = 3x, then du/dx = 3. Applying the chain rule, we get:
d/dx (sin(u)) = cos(u) * du/dx = cos(3x) * 3 = 3cos(3x).
Now, we can rewrite the limit as the limit of the derivatives:
lim x→0 (7x)/(sin(3x)) = lim x→0 7/(3cos(3x)).
Now, we can directly evaluate the limit as x approaches 0. Plugging in x=0 to the new expression, we get:
lim x→0 7/(3cos(3x)) = 7/(3cos(0)) = 7/(3*1) = 7/3.
Therefore, the limit of the function (7x)/(sin(3x)) as x approaches 0 is 7/3.
More Answers:
[next_post_link]