Evaluate the following limit using L'Hospital's rule. $\newline$ $\lim _{x \rightarrow 0^{+}}(\cos (9 x))\left(\frac{1}{9 x^{2}}\right)$

View step-by-step help

Home

Math Problems

Calculus

Find limits involving trigonometric functions

Full solution

Q. Evaluate the following limit using L'Hospital's rule.

\newline

\lim _{x \rightarrow 0^{+}}(\cos (9 x))\left(\frac{1}{9 x^{2}}\right)

Apply L'Hospital's Rule: Use L'Hospital's Rule since we have an indeterminate form of $0/0$ . Differentiate the numerator and the denominator separately. Numerator derivative: $\frac{d}{dx}[\cos(9x)] = -9\sin(9x)$ Denominator derivative: $\frac{d}{dx}[9x^2] = 18x$
Differentiate and Simplify: Apply L'Hospital's Rule by substituting the derivatives. $\newline$ $\lim_{x \to 0^{+}}\left(\frac{-9\sin(9x)}{18x}\right)$ $\newline$ Simplify the expression. $\newline$ \lim_{x \to $0$^{+}}\left(\frac{$-9$\sin($9$x)}{$18$x}\right) = \lim_{x \to $0$^{+}}\left(\frac{-\sin($9$x)}{$2$x}\right)
Evaluate Limit: Evaluate the limit of the simplified expression as \(x approaches $0$ . $\lim_{x \rightarrow 0^{+}}\left(-\frac{\sin(9x)}{2x}\right) = \frac{-\sin(0)}{2\cdot 0}$ This results in an indeterminate form again, so we need to apply L'Hospital's Rule once more.
Apply L'Hospital's Rule: Differentiate the numerator and the denominator again. $\newline$ Numerator derivative: $\frac{d}{dx}[-\sin(9x)] = -9\cos(9x)$ $\newline$ Denominator derivative: $\frac{d}{dx}[2x] = 2$
Differentiate and Evaluate: Apply L'Hospital's Rule by substituting the new derivatives. $\newline$ $\lim_{x \to 0^{+}}\left(-\frac{9\cos(9x)}{2}\right)$ $\newline$ Evaluate the limit of the new expression as x approaches $0$ . $\newline$ $\lim_{x \to 0^{+}}\left(-\frac{9\cos(9x)}{2}\right) = \left(-\frac{9\cos(0)}{2}\right)$
Calculate Final Value: Calculate the final value. $\newline$ $(-9\cos(0))/2 = (-9\times1)/2$ $\newline$ $= -\frac{9}{2}$