Question

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

Short answer

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

Step-by-step solution

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.

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, \(\lim _{x \rightarrow 0} \frac{a-\cos bx}{x^{2}}\) = \(\lim _{x \rightarrow 0} \frac{0 + bsin(bx)}{2x}\).

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 \(b^2/2\).

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 \(b = ±2\).

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 \(a = 0\).