Use L'Hôpital's Rule (possibly more than once) to evaluate the following limit. (Use symbolic notation and fractions where needed.) $\newline$ $\lim_{x \to 0}\frac{\sin(7x)-7x \cos(7x)}{7x-\sin(7x)}=$

View step-by-step help

Home

Math Problems

Algebra 1

Evaluate piecewise-defined functions

Full solution

Q. Use L'Hôpital's Rule (possibly more than once) to evaluate the following limit. (Use symbolic notation and fractions where needed.)

\newline

\lim_{x \to 0}\frac{\sin(7x)-7x \cos(7x)}{7x-\sin(7x)}=

Check Indeterminate Form: We are given the limit to evaluate using L'Hôpital's Rule: $\newline$ $\lim_{x \to 0} \frac{\sin(7x) - 7x \cos(7x)}{7x - \sin(7x)}$ $\newline$ First, we need to check if the limit is in an indeterminate form that allows us to apply L'Hôpital's Rule.
Evaluate Numerator and Denominator: Evaluate the limit of the numerator and the denominator separately as $x$ approaches $0$ : $\lim_{x \to 0} \sin(7x) = 0$ $\lim_{x \to 0} 7x \cos(7x) = 0$ $\lim_{x \to 0} 7x = 0$ $\lim_{x \to 0} \sin(7x) = 0$ Both the numerator and the denominator approach $0$ , so we have an indeterminate form of $0/0$ .
Apply L'Hôpital's Rule: Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that if $\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{0}{0}$ or $\pm\infty/\pm\infty$ , then $\lim_{x \to c} \frac{f(x)}{g(x)}$ can be found by $\lim_{x \to c} \frac{f'(x)}{g'(x)}$ , provided the latter limit exists.
Find Derivatives: We will find the derivatives of the numerator and the denominator: $\newline$ The derivative of the numerator $f(x) = \sin(7x) - 7x \cos(7x)$ is $f'(x) = 7 \cos(7x) - 7 \cos(7x) + 49x \sin(7x)$ . $\newline$ The derivative of the denominator $g(x) = 7x - \sin(7x)$ is $g'(x) = 7 - 7 \cos(7x)$ .
Simplify Derivative: Simplify the derivative of the numerator: $f'(x) = 7 \cos(7x) - 7 \cos(7x) + 49x \sin(7x)$ simplifies to $f'(x) = 49x \sin(7x)$ .
Apply Rule Again: Now we apply L'Hôpital's Rule by taking the limit of the derivatives: $\lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{49x \sin(7x)}{7 - 7 \cos(7x)}$
Find New Derivatives: Evaluate the limit of the new expression as $x$ approaches $0$ :
$\lim_{x \to 0} (49x \sin(7x)) = 0$
$\lim_{x \to 0} (7 - 7 \cos(7x)) = 0$
We again have an indeterminate form of $0/0$ , so we need to apply L'Hôpital's Rule once more.
Evaluate New Expression: Find the derivatives of the new numerator and denominator: $\newline$ The derivative of the new numerator $h(x) = 49x \sin(7x)$ is $h'(x) = 49 \sin(7x) + 343x \cos(7x)$ . $\newline$ The derivative of the new denominator $k(x) = 7 - 7 \cos(7x)$ is $k'(x) = 7 \sin(7x)$ .
Re-evaluate Derivatives: Apply L'Hôpital's Rule again by taking the limit of the new derivatives: $\lim_{x \to 0} \frac{h'(x)}{k'(x)} = \lim_{x \to 0} \frac{(49 \sin(7x) + 343x \cos(7x))}{(7 \sin(7x))}$
Re-evaluate Derivatives: Apply L'Hôpital's Rule again by taking the limit of the new derivatives: $\newline$ $\lim_{x \to 0} \frac{h'(x)}{k'(x)} = \lim_{x \to 0} \frac{49 \sin(7x) + 343x \cos(7x)}{7 \sin(7x)}$ Evaluate the limit of the new expression as x approaches $0$ : $\newline$ $\lim_{x \to 0} (49 \sin(7x) + 343x \cos(7x)) = 49 \times 0 + 343 \times 0 \times 1 = 0$ $\newline$ $\lim_{x \to 0} (7 \sin(7x)) = 7 \times 0 = 0$ $\newline$ We still have an indeterminate form of $0/0$ , which means we made a mistake in our differentiation or simplification. We need to re-evaluate the derivatives.