For what values of a and b is the following equation true?
( (sin2x)/ x^3 + a + b/(x^2))
lim x--->0
a=
b=
I don't see an equation.
However, sin2x/x^3 = (sin2x/(2x)) * 2/x^2
and you know limit of sunu/u = 1
To find the values of a and b for which the equation holds true, we need to compute the limit of the expression as x approaches 0.
The given equation is:
((sin2x)/x^3 + a + b/(x^2))
To find the limit of this expression as x approaches 0, we can simplify it by factoring out common terms in the numerator and denominator.
Let's start by factoring out x^3 from the first term in the numerator:
((sin2x)/x^3) = (sin2x * (1/x^3))
Next, let's factor out x^2 from the second term in the numerator:
(b/(x^2)) = (b * (1/x^2))
Now, we have the simplified form of the expression:
(1/x^3) * (sin2x) + a + (1/x^2) * b
To evaluate the limit, substitute x = 0 into the simplified expression:
lim x--> 0 (1/x^3) * (sin2x) + a + (1/x^2) * b
When x approaches 0, the first two terms contain factors of 1/x^3 and 1/x^2, respectively. These factors tend towards infinity as x approaches 0, and since the expression is being multiplied by them, the overall limit will be affected by their behavior.
For the limit to exist, the terms with 1/x^3 and 1/x^2 should cancel each other out. This means their coefficients a and b should be chosen such that:
a = -1
b = 0
With these values of a and b, the expression becomes:
(1/x^3) * (sin2x) - 1
Now, let's evaluate the limit as x approaches 0 with a = -1 and b = 0:
lim x--> 0 (1/x^3) * (sin2x) - 1
The limit of (1/x^3) * (sin2x) as x approaches 0 can be determined using L'Hopital's rule or by recognizing that (sin2x)/x^3 is equivalent to 2/x, and the limit as x approaches 0 is 2/0, which tends towards infinity. Therefore, the overall limit as x approaches 0 with a = -1 and b = 0 is -∞.
So, for the equation to hold true, a must be -1 and b must be 0.