Problem 75 - Essential Calculus

Question

Find the values of \(a\) and \(b\) such that \(\lim _{x \rightarrow 0} \frac{a-\cos b x}{x^{2}}=2\).

Step-by-Step Solution

Answer

The values of \(a\) and \(b\) are 0 and ±2 respectively.

Step 1: Analyze the equation

The given equation is \(\lim _{x \rightarrow 0} \frac{a-\cos bx}{x^{2}}=2\). This scenario is an indeterminate form and it is required to determine the values of \(a\) and \(b\). L'Hopital's rule will need to be used to solve the problem.

Step 2: Apply L'Hopital's Rule

According to L'Hopital’s rule, when the limit of a function as \(x\) approaches a certain value is indeterminate (i.e., either \(0/0\) or \(\infty/ \infty)\), the required limit can be found by computing the limit of the derivative of the numerator and denominator. The derivative of \(a\) is 0 as it's a constant, and the derivative of \(-cos(bx)\) is \(bsin(bx)\), while the derivative of \(x^2\) is \(2x\). Thus, \(\lim _{x \rightarrow 0} \frac{a-\cos bx}{x^{2}}\) = \(\lim _{x \rightarrow 0} \frac{0 + bsin(bx)}{2x}\).

Step 3: Simplify the equation

We again use L'Hopital's rule to find the limit as x goes to 0 for the equation \(\lim _{x \rightarrow 0} \frac{bsin(bx)}{2x}\). The derivative of \(bsin(bx)\) is \(b^2cos(bx)\) and derivative of \(2x\) is 2. Thus, the equation simplifies to \(\lim _{x \rightarrow 0} \frac{b^2cos(bx)}{2}\). This gives the result as \(b^2/2\).

Step 4: Equate to 2

Since we know from step 3 that our limit returns \(b^2/2\), we can equate this to 2 (the result of our limit as defined in the problem). Solve the equation \(b^2/2 = 2\). Solve for \(b\) to get \(b = ±2\).

Step 5: Find the Value of a

Substitute the value of \(b\) into \(\lim _{x \rightarrow 0} \frac{a-\cos bx}{x^{2}}=2\) to find the value of \(a\). Since the \(b\)\(-cos(bx)\) term will go to 0 as \(x\) approaches 0, the equation can be rewritten as follows: \(\lim _{x \rightarrow 0} \frac{a}{x^{2}}=2\). Simplifying this, we find that \(a = 0\).