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.

23.\(\mathop {\lim }\limits_{x \to 0} \cot 2x\sin 6x\)

Short answer

The value of \(\mathop {\lim }\limits_{x \to 0} \cot 2x\sin 6x\) is .

Step-by-step solution

Given information.

The given limit is \(\mathop {\lim }\limits_{x \to 0} \cot 2x\sin 6x\).

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\( - \infty \)).

Find the value of the limit

Simplify the given limit.

\(\begin{array}{c}\mathop {\lim }\limits_{x \to 0} \cot 2x\sin 6x = \mathop {\lim }\limits_{x \to 0} \frac{{\cos 2x}}{{\sin 2x}}\sin 6x\\ = \mathop {\lim }\limits_{x \to 0} \frac{{\cos 2x\sin 6x}}{{\sin 2x}}\end{array}\)

Here\(f(x) = \cos 2x\sin 6x\)and\(g(x) = \sin 2x\).

Evaluate the limit of the numerator.

\(\begin{array}{c}\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} \cos 2x\sin 6x\\ = 1 \cdot 0\\ = 0\end{array}\)

Evaluate the limit of the denominator.

\(\begin{array}{c}\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} \sin 2x\\ = 0\end{array}\)

Since\(\mathop {\lim }\limits_{x \to 0} f(x) = 0\)and\(\mathop {\lim }\limits_{x \to 0} g(x) = 0\), L'Hospital's rule can be applied.

Apply L’Hospital’s rule

Apply L'Hospital's rule and simplify the given limit.

\(\begin{array}{c}\mathop {\lim }\limits_{x \to 0} \frac{{\cos 2x\sin 6x}}{{\sin 2x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}(\cos 2x\sin 6x)}}{{\frac{d}{{dx}}(\sin 2x)}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{\cos 2x\cos 6x(6) + \sin 2x(2)\sin 6x}}{{\cos 2x(2)}}\\ = \frac{6}{2}\\ = 3\end{array}\)

Therefore, the value of \(\mathop {\lim }\limits_{x \to 1} \frac{{{x^a} - ax + a - 1}}{{{{(x - 1)}^2}}}\) is .