lim f(x) sin 3x / x
x�¨0
You can use the following Taylor series expansion for sin(3x):
sin 3x = 3x -(1/2)(3x)^2 + (1/3!)(3x)^3 -...
Divide that by x and you get a limit of 3.
You could also use L'Hopital's rule.
To find the limit of a function as x approaches 0, we can try direct substitution. However, in this case, if we directly substitute x = 0 into the function, we get an indeterminate form of the type 0/0, which does not give us a clear result.
To solve this problem, we can use a common limit rule known as L'Hospital's Rule. This rule states that if we have an indeterminate form of the type 0/0 or ∞/∞, we can take the derivative of the numerator and denominator separately and then evaluate the limit again.
Start by taking the derivative of the numerator and denominator separately. Recall that the derivative of sin x is cos x. Applying the chain rule, the derivative of sin(3x) with respect to x is 3cos(3x). The derivative of x with respect to x is simply 1.
Now, we have:
lim(x->0) (f(x)sin(3x))/x = lim(x->0) (f(x)*3cos(3x))/(1)
Next, evaluate the limit as x approaches 0. This time, we can substitute x = 0 directly into the function, as we now have a non-indeterminate form:
lim(x->0) (f(x)*3cos(3x))/(1) = f(0) * 3cos(0)/1
Since cos(0) = 1 and f(0) is unknown based on the given information, the final answer is:
lim(x->0) (f(x)sin(3x))/x = 3f(0)