Question

1–38 ■ Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply, explain why.

16. \(\mathop {\lim }\limits_{x \to 0} \frac{{\cos mx - \cos nx}}{{{x^2}}}\).

Short answer

The value of \(\mathop {\lim }\limits_{x \to 0} \frac{{\cos mx - \cos nx}}{{{x^2}}}\)is\(\frac{{{n^2} - {m^2}}}{2}\).

Step-by-step solution

Given limit

The given limit is\(\mathop {\lim }\limits_{x \to 0} \frac{{\cos mx - \cos nx}}{{{x^2}}}\).

Concept of L'Hospital's Rule

Suppose\(f\)and\(g\)are differentiable and\({g^\prime }(x) \ne 0\)near\(a\)(except possibly at\(a\)).

Suppose that\(\mathop {\lim }\limits_{x \to a} f(x) = 0\)and\(\mathop {\lim }\limits_{x \to a} g(x) = 0\)or that\(\mathop {\lim }\limits_{x \to a} f(x) = \pm \infty \)and\(\mathop {\lim }\limits_{x \to a} g(x) = \pm \infty \)or in other words, we have an indeterminate form of type\(\frac{0}{0}\)or\(\frac{\infty }{\infty }\).

Then,\(\mathop {\lim }\limits_{x \to a} \frac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to a} \frac{{{f^\prime }(x)}}{{{g^\prime }(x)}}\)if the limit on the right side exists (or is\(\infty \)or\( - \infty \)).

Evaluate the given limit

Obtain the value of the function as \(x\)approaches \(0\).

As \(x\) approaches 0, the numerator is:

\(\begin{array}{c}\cos mx - \cos nx = \cos m \cdot 0 - \cos n \cdot 0\\ = \cos 0 - \cos 0\\ = 1 - 1\\ = 0\end{array}\)

And the denominator \({x^2}\) approaches to \(0\).

Thus, \(\mathop {\lim }\limits_{x \to 0} \frac{{\cos mx - \cos nx}}{{{x^2}}} = \frac{0}{0}\) is in an indeterminate form.

Therefore, apply L'Hospital's Rule and obtain the limit.

\(\mathop {\lim }\limits_{x \to 0} \frac{{\cos mx - \cos nx}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{ - m\sin mx + n\sin nx}}{{2x}}\)

Again as \(x\) approaches 0, the numerator is:

\(\begin{array}{c} - m\sin mx + n\sin nx = - m\sin m \cdot 0 + n\sin n \cdot 0\\ = - m \cdot 0 + n \cdot 0\\ = 0 + 0\\ = 0\end{array}\)

And as \(x\) approaches 0, the denominator 2 x approaches to \(0\).

Thus, \(\mathop {\lim }\limits_{x \to 0} \frac{{ - m\sin mx + n\sin nx}}{{2x}} = \frac{0}{0}\) is again in an indeterminate form.

Apply L'Hospital's Rule again and obtain the limit as.

\(\begin{array}{c}\mathop {\lim }\limits_{x \to 0} \frac{{ - m\sin mx + n\sin nx}}{{2x}} = \mathop {\lim }\limits_{x \to 0} \frac{{ - {m^2}\cos mx + {n^2}\cos nx}}{2}\\ = \frac{{ - {m^2}\cos m.0 + {n^2}\cos n \cdot 0}}{2}\\ = \frac{{ - {m^2} \cdot 1 + {n^2} \cdot 1}}{2}\\ = \frac{{{n^2} - {m^2}}}{2}\end{array}\)

Thus, the limit function is\(\frac{{{n^2} - {m^2}}}{2}\).