AP Calculus AB L'Hospital's Rule Worksheet Evaluate the limits: 1. \( \lim _{x \rightarrow 4} \frac{3 x-12}{x^{2}-16} \) 2. \( \lim _{x \rightarrow-4} \frac{2 x^{2}+13 x+20}{x+4} \) 3. \( \lim _{x \rightarrow 6} \frac{\sqrt{x+10}-4}{x-6} \) 4. \( \lim _{x \rightarrow 0} \frac{\sin 6 x}{4 x} \) 5. \( \lim _{x \rightarrow 3} \frac{x^{2}-2 x-3}{x-3} \) 6. \( \lim _{x \rightarrow 0} \frac{\sqrt{25-x^{2}}-5}{x} \) 7. \( \lim _{x \rightarrow 0} \frac{\sin 3 x}{\sin 5 x} \) 8. \( \lim _{h \rightarrow 0} \frac{\sin (x+h)-\sin x}{h} \) 9. \( \lim _{x \rightarrow \infty} \frac{5 x^{2}+3 x-1}{4 x^{2}+5} \) 10. \( \lim _{x \rightarrow 0} \frac{\sin ^{2} 2 x}{5 x^{2}} \) 11. \( \lim _{x \rightarrow 0} \frac{\sin 5 x}{\tan 9 x} \) 12. \( \lim _{x \rightarrow 5} \frac{\sqrt{x+4}-3}{\sqrt{x-1}-2} \) 13. \( \lim _{x \rightarrow \frac{\pi}{3}} \frac{\cos x-\frac{1}{2}}{x-\frac{\pi}{3}} \) 14. \( \lim _{x \rightarrow 3} \frac{\frac{1}{x}-\frac{1}{3}}{x^{2}-9} \) 15. \( \lim _{x \rightarrow 2} \frac{x^{2}-4 x+4}{x^{2}-x-2} \)

Solve the quadratic equation by following steps: - step0: Solve by factoring: \(x^{2}-2x-3=0\) - step1: Factor the expression: \(\left(x-3\right)\left(x+1\right)=0\) - step2: Separate into possible cases: \(\begin{align}&x-3=0\\&x+1=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=3\\&x=-1\end{align}\) - step4: Rewrite: \(x_{1}=-1,x_{2}=3\) Solve the equation \( \frac{x^{2}-4x+4}{x^{2}-x-2}=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{x^{2}-4x+4}{x^{2}-x-2}=0\) - step1: Find the domain: \(\frac{x^{2}-4x+4}{x^{2}-x-2}=0,x \in \left(-\infty,-1\right)\cup \left(-1,2\right)\cup \left(2,+\infty\right)\) - step2: Divide the terms: \(\frac{x-2}{x+1}=0\) - step3: Cross multiply: \(x-2=\left(x+1\right)\times 0\) - step4: Simplify the equation: \(x-2=0\) - step5: Move the constant to the right side: \(x=0+2\) - step6: Remove 0: \(x=2\) - step7: Check if the solution is in the defined range: \(x=2,x \in \left(-\infty,-1\right)\cup \left(-1,2\right)\cup \left(2,+\infty\right)\) - step8: Find the intersection: \(x \in \varnothing \) Solve the equation \( 2x^{2}+13x+20=0 \). Solve the quadratic equation by following steps: - step0: Solve by factoring: \(2x^{2}+13x+20=0\) - step1: Factor the expression: \(\left(x+4\right)\left(2x+5\right)=0\) - step2: Separate into possible cases: \(\begin{align}&2x+5=0\\&x+4=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=-\frac{5}{2}\\&x=-4\end{align}\) - step4: Rewrite: \(x_{1}=-4,x_{2}=-\frac{5}{2}\) Solve the equation \( \frac{\sin^{2}(2x)}{5x^{2}}=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{\sin^{2}\left(2x\right)}{5x^{2}}=0\) - step1: Find the domain: \(\frac{\sin^{2}\left(2x\right)}{5x^{2}}=0,x\neq 0\) - step2: Cross multiply: \(\sin^{2}\left(2x\right)=5x^{2}\times 0\) - step3: Simplify the equation: \(\sin^{2}\left(2x\right)=0\) - step4: Set the base equal to 0: \(\sin\left(2x\right)=0\) - step5: Use the inverse trigonometric function: \(2x=\arcsin\left(0\right)\) - step6: Calculate: \(2x=0\) - step7: Add the period: \(2x=k\pi ,k \in \mathbb{Z}\) - step8: Solve the equation: \(x=\frac{k\pi }{2},k \in \mathbb{Z}\) - step9: Check if the solution is in the defined range: \(x=\frac{k\pi }{2},k \in \mathbb{Z},x\neq 0\) - step10: Find the intersection: \(x=\frac{k\pi }{2},k \in \left(-\infty,0\right),k \in \mathbb{Z}\cup k \in \left(0,+\infty\right),k \in \mathbb{Z}\) Solve the equation \( \frac{\sqrt(25-x^{2})-5}{x}=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{\sqrt{25-x^{2}}-5}{x}=0\) - step1: Find the domain: \(\frac{\sqrt{25-x^{2}}-5}{x}=0,x \in \left[-5,0\right)\cup \left(0,5\right]\) - step2: Cross multiply: \(\sqrt{25-x^{2}}-5=x\times 0\) - step3: Simplify the equation: \(\sqrt{25-x^{2}}-5=0\) - step4: Move the constant to the right-hand side: \(\sqrt{25-x^{2}}=5\) - step5: Raise both sides to the \(2\)-th power\(:\) \(\left(\sqrt{25-x^{2}}\right)^{2}=5^{2}\) - step6: Evaluate the power: \(25-x^{2}=25\) - step7: Move the constant to the right side: \(-x^{2}=25-25\) - step8: Subtract the terms: \(-x^{2}=0\) - step9: Change the signs: \(x^{2}=0\) - step10: Set the base equal to 0: \(x=0\) - step11: Check if the solution is in the defined range: \(x=0,x \in \left[-5,0\right)\cup \left(0,5\right]\) - step12: Find the intersection: \(x \in \varnothing \) Solve the equation \( \frac{\sin(3x)}{\sin(5x)}=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{\sin\left(3x\right)}{\sin\left(5x\right)}=0\) - step1: Find the domain: \(\frac{\sin\left(3x\right)}{\sin\left(5x\right)}=0,x\neq \frac{k\pi }{5},k \in \mathbb{Z}\) - step2: Cross multiply: \(\sin\left(3x\right)=\sin\left(5x\right)\times 0\) - step3: Simplify the equation: \(\sin\left(3x\right)=0\) - step4: Use the inverse trigonometric function: \(3x=\arcsin\left(0\right)\) - step5: Calculate: \(3x=0\) - step6: Add the period: \(3x=k\pi ,k \in \mathbb{Z}\) - step7: Solve the equation: \(x=\frac{k\pi }{3},k \in \mathbb{Z}\) - step8: Check if the solution is in the defined range: \(x=\frac{k\pi }{3},k \in \mathbb{Z},x\neq \frac{k\pi }{5},k \in \mathbb{Z}\) - step9: Find the intersection: \(x=\left\{ \begin{array}{l}\frac{\pi }{3}+k\pi \\\frac{2\pi }{3}+k\pi \end{array}\right.,k \in \mathbb{Z}\) Solve the equation \( \frac{5x^{2}+3x-1}{4x^{2}+5}=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{5x^{2}+3x-1}{4x^{2}+5}=0\) - step1: Cross multiply: \(5x^{2}+3x-1=\left(4x^{2}+5\right)\times 0\) - step2: Simplify the equation: \(5x^{2}+3x-1=0\) - step3: Solve using the quadratic formula: \(x=\frac{-3\pm \sqrt{3^{2}-4\times 5\left(-1\right)}}{2\times 5}\) - step4: Simplify the expression: \(x=\frac{-3\pm \sqrt{3^{2}-4\times 5\left(-1\right)}}{10}\) - step5: Simplify the expression: \(x=\frac{-3\pm \sqrt{29}}{10}\) - step6: Separate into possible cases: \(\begin{align}&x=\frac{-3+\sqrt{29}}{10}\\&x=\frac{-3-\sqrt{29}}{10}\end{align}\) - step7: Rewrite the fraction: \(\begin{align}&x=\frac{-3+\sqrt{29}}{10}\\&x=-\frac{3+\sqrt{29}}{10}\end{align}\) - step8: Rewrite: \(x_{1}=-\frac{3+\sqrt{29}}{10},x_{2}=\frac{-3+\sqrt{29}}{10}\) Solve the equation \( \frac{\sin(5x)}{\tan(9x)}=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{\sin\left(5x\right)}{\tan\left(9x\right)}=0\) - step1: Find the domain: \(\frac{\sin\left(5x\right)}{\tan\left(9x\right)}=0,x\neq \frac{k\pi }{18},k \in \mathbb{Z}\) - step2: Cross multiply: \(\sin\left(5x\right)=\tan\left(9x\right)\times 0\) - step3: Simplify the equation: \(\sin\left(5x\right)=0\) - step4: Use the inverse trigonometric function: \(5x=\arcsin\left(0\right)\) - step5: Calculate: \(5x=0\) - step6: Add the period: \(5x=k\pi ,k \in \mathbb{Z}\) - step7: Solve the equation: \(x=\frac{k\pi }{5},k \in \mathbb{Z}\) - step8: Check if the solution is in the defined range: \(x=\frac{k\pi }{5},k \in \mathbb{Z},x\neq \frac{k\pi }{18},k \in \mathbb{Z}\) - step9: Find the intersection: \(x=\left\{ \begin{array}{l}\frac{\pi }{5}+k\pi \\\frac{2\pi }{5}+k\pi \\\frac{3\pi }{5}+k\pi \\\frac{4\pi }{5}+k\pi \end{array}\right.,k \in \mathbb{Z}\) Solve the equation \( \frac{\sqrt(x+10)-4}{x-6}=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{\sqrt{x+10}-4}{x-6}=0\) - step1: Find the domain: \(\frac{\sqrt{x+10}-4}{x-6}=0,x \in \left[-10,6\right)\cup \left(6,+\infty\right)\) - step2: Cross multiply: \(\sqrt{x+10}-4=\left(x-6\right)\times 0\) - step3: Simplify the equation: \(\sqrt{x+10}-4=0\) - step4: Move the constant to the right-hand side: \(\sqrt{x+10}=4\) - step5: Raise both sides to the \(2\)-th power\(:\) \(\left(\sqrt{x+10}\right)^{2}=4^{2}\) - step6: Evaluate the power: \(x+10=16\) - step7: Move the constant to the right side: \(x=16-10\) - step8: Subtract the numbers: \(x=6\) - step9: Check if the solution is in the defined range: \(x=6,x \in \left[-10,6\right)\cup \left(6,+\infty\right)\) - step10: Find the intersection: \(x \in \varnothing \) Solve the equation \( \frac{\sin(6x)}{4x}=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{\sin\left(6x\right)}{4x}=0\) - step1: Find the domain: \(\frac{\sin\left(6x\right)}{4x}=0,x\neq 0\) - step2: Cross multiply: \(\sin\left(6x\right)=4x\times 0\) - step3: Simplify the equation: \(\sin\left(6x\right)=0\) - step4: Use the inverse trigonometric function: \(6x=\arcsin\left(0\right)\) - step5: Calculate: \(6x=0\) - step6: Add the period: \(6x=k\pi ,k \in \mathbb{Z}\) - step7: Solve the equation: \(x=\frac{k\pi }{6},k \in \mathbb{Z}\) - step8: Check if the solution is in the defined range: \(x=\frac{k\pi }{6},k \in \mathbb{Z},x\neq 0\) - step9: Find the intersection: \(x=\frac{k\pi }{6},k \in \left(-\infty,0\right),k \in \mathbb{Z}\cup k \in \left(0,+\infty\right),k \in \mathbb{Z}\) Solve the equation \( \frac{\sqrt(x+4)-3}{\sqrt(x-1)-2}=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{\sqrt{x+4}-3}{\sqrt{x-1}-2}=0\) - step1: Find the domain: \(\frac{\sqrt{x+4}-3}{\sqrt{x-1}-2}=0,x \in \left[1,5\right)\cup \left(5,+\infty\right)\) - step2: Cross multiply: \(\sqrt{x+4}-3=\left(\sqrt{x-1}-2\right)\times 0\) - step3: Simplify the equation: \(\sqrt{x+4}-3=0\) - step4: Move the constant to the right-hand side: \(\sqrt{x+4}=3\) - step5: Raise both sides to the \(2\)-th power\(:\) \(\left(\sqrt{x+4}\right)^{2}=3^{2}\) - step6: Evaluate the power: \(x+4=9\) - step7: Move the constant to the right side: \(x=9-4\) - step8: Subtract the numbers: \(x=5\) - step9: Check if the solution is in the defined range: \(x=5,x \in \left[1,5\right)\cup \left(5,+\infty\right)\) - step10: Find the intersection: \(x \in \varnothing \) Solve the equation \( 3x-12=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(3x-12=0\) - step1: Move the constant to the right side: \(3x=0+12\) - step2: Remove 0: \(3x=12\) - step3: Divide both sides: \(\frac{3x}{3}=\frac{12}{3}\) - step4: Divide the numbers: \(x=4\) Solve the equation \( \frac{\frac{1}{x}-\frac{1}{3}}{x^{2}-9}=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{\frac{1}{x}-\frac{1}{3}}{x^{2}-9}=0\) - step1: Divide the terms: \(-\frac{1}{3x\left(x+3\right)}=0\) - step2: Rewrite the expression: \(\frac{-1}{3x\left(x+3\right)}=0\) - step3: Cross multiply: \(-1=3x\left(x+3\right)\times 0\) - step4: Simplify the equation: \(-1=0\) - step5: The statement is false: \(x \in \varnothing \) Solve the equation \( \frac{\cos(x)-\frac{1}{2}}{x-\frac{\pi}{3}}=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{\cos\left(x\right)-\frac{1}{2}}{x-\frac{\pi }{3}}=0\) - step1: Find the domain: \(\frac{\cos\left(x\right)-\frac{1}{2}}{x-\frac{\pi }{3}}=0,x\neq \frac{\pi }{3}\) - step2: Divide the terms: \(\frac{6\cos\left(x\right)-3}{2\left(3x-\pi \right)}=0\) - step3: Cross multiply: \(6\cos\left(x\right)-3=2\left(3x-\pi \right)\times 0\) - step4: Simplify the equation: \(6\cos\left(x\right)-3=0\) - step5: Move the constant to the right side: \(6\cos\left(x\right)=0+3\) - step6: Remove 0: \(6\cos\left(x\right)=3\) - step7: Divide both sides: \(\frac{6\cos\left(x\right)}{6}=\frac{3}{6}\) - step8: Divide the numbers: \(\cos\left(x\right)=\frac{1}{2}\) - step9: Use the inverse trigonometric function: \(x=\arccos\left(\frac{1}{2}\right)\) - step10: Calculate: \(\begin{align}&x=\frac{\pi }{3}\\&x=\frac{5\pi }{3}\end{align}\) - step11: Add the period: \(\begin{align}&x=\frac{\pi }{3}+2k\pi ,k \in \mathbb{Z}\\&x=\frac{5\pi }{3}+2k\pi ,k \in \mathbb{Z}\end{align}\) - step12: Find the union: \(x=\left\{ \begin{array}{l}\frac{\pi }{3}+2k\pi \\\frac{5\pi }{3}+2k\pi \end{array}\right.,k \in \mathbb{Z}\) - step13: Check if the solution is in the defined range: \(x=\left\{ \begin{array}{l}\frac{\pi }{3}+2k\pi \\\frac{5\pi }{3}+2k\pi \end{array}\right.,k \in \mathbb{Z},x\neq \frac{\pi }{3}\) - step14: Find the intersection: \(x=\left\{ \begin{array}{l}\frac{\pi }{3}+2k\pi ,k \in \left(-\infty,0\right),k \in \mathbb{Z}\cup k \in \left(0,+\infty\right),k \in \mathbb{Z}\\\frac{5\pi }{3}+2k\pi ,k \in \left(-\frac{2}{3},+\infty\right),k \in \mathbb{Z}\cup k \in \left(-\infty,-\frac{2}{3}\right),k \in \mathbb{Z}\end{array}\right.\) Let's evaluate the limits step by step using L'Hospital's Rule where applicable. ### 1. \( \lim _{x \rightarrow 4} \frac{3 x-12}{x^{2}-16} \) First, substitute \( x = 4 \): \[ \frac{3(4) - 12}{4^2 - 16} = \frac{0}{0} \] This is an indeterminate form, so we apply L'Hospital's Rule: \[ \lim_{x \to 4} \frac{3}{2x} = \frac{3}{2(4)} = \frac{3}{8} \] ### 2. \( \lim _{x \rightarrow -4} \frac{2 x^{2}+13 x+20}{x+4} \) Substituting \( x = -4 \): \[ \frac{2(-4)^2 + 13(-4) + 20}{-4 + 4} = \frac{32 - 52 + 20}{0} = \frac{0}{0} \] Applying L'Hospital's Rule: \[ \lim_{x \to -4} \frac{4x + 13}{1} = 4(-4) + 13 = -16 + 13 = -3 \] ### 3. \( \lim _{x \rightarrow 6} \frac{\sqrt{x+10}-4}{x-6} \) Substituting \( x = 6 \): \[ \frac{\sqrt{6+10} - 4}{6-6} = \frac{0}{0} \] Applying L'Hospital's Rule: \[ \lim_{x \to 6} \frac{\frac{1}{2\sqrt{x+10}}}{1} = \frac{1}{2\sqrt{16}} = \frac{1}{8} \] ### 4. \( \lim _{x \rightarrow 0} \frac{\sin 6 x}{4 x} \) Substituting \( x = 0 \): \[ \frac{\sin(0)}{0} = \frac{0}{0} \] Applying L'Hospital's Rule: \[ \lim_{x \to 0} \frac{6\cos(6x)}{4} = \frac{6\cos(0)}{4} = \frac{6}{4} = \frac{3}{2} \] ### 5. \( \lim _{x \rightarrow 3} \frac{x^{2}-2 x-3}{x-3} \) Substituting \( x = 3 \): \[ \frac{3^2 - 2(3) - 3}{3-3} = \frac{0}{0} \] Applying L'Hospital's Rule: \[ \lim_{x \to 3} \frac{2x - 2}{1} = 2(3) - 2 = 6 - 2 = 4 \] ### 6. \( \lim _{x \rightarrow 0} \frac{\sqrt{25-x^{2}}-5}{x} \) Substituting \( x = 0 \): \[ \frac{\sqrt{25-0}-5}{0} = \frac{0}{0} \] Applying L'Hospital's Rule: \[ \lim_{x \to 0} \frac{-\frac{x}{\sqrt{25-x^2}}}{1} = -\frac{0}{5} = 0 \] ### 7. \( \lim _{x \rightarrow 0} \frac{\sin 3 x}{\sin 5 x} \) Substituting \( x = 0 \): \[ \frac{\sin(0)}{\sin(0)} = \frac{0}{0} \] Applying L'Hospital's Rule: \[ \lim_{x \to 0} \frac{3\cos(3x)}{5\cos(5x)} = \frac{3}{5} \] ### 8. \( \lim _{h \rightarrow 0} \frac{\sin (x+h)-\sin x}{h} \) This is the definition of the derivative of \( \sin x \): \[ \lim_{h \to 0} \frac{\sin (x+h)-\sin x}{h} = \cos x \] ### 9. \( \lim _{x \rightarrow \infty} \frac{5 x^{2}+3 x-1}{4 x^{2}+5} \) Dividing numerator and denominator by \( x^2 \): \[ \lim_{x \to \infty} \frac{5 + \frac{3}{x} - \frac{1}{x^2}}{4 + \frac{5}{x^2}} = \frac{5}{4} \] ### 10. \( \lim _{x \rightarrow 0} \frac{\sin ^{2} 2 x}{5 x^{2}} \) Substituting \( x = 0 \): \[ \frac{0}{0} \] Applying L'Hospital's Rule: \[ \lim_{x \to 0} \frac{2\sin(2x)\cdot 2\cos(2x)}{10x} = \frac{4\cdot 1}{10} = \frac{2}{5} \] ### 11. \( \lim _{x \rightarrow 0} \frac{\sin 5 x}{\tan 9 x} \) Substituting \( x = 0 \): \[ \frac{0}{0} \] Applying L'Hospital's Rule: \[ \lim_{x \to 0} \frac{5\cos(5x)}{9\sec^2(9x)} = \frac{5}{9} \] ### 12. \( \lim _{x \rightarrow 5} \frac{\sqrt{x+4}-3}{\sqrt{x-1}-2} \) Substituting \( x = 5 \): \[ \frac{\sqrt{5+4}-3}{\sqrt{5-1}-2} = \frac{3-3}{2-2} = \frac{0}{0} \] Applying L'Hospital's Rule: \[ \lim_{x \to 5} \frac{\frac{1}{2\sqrt{x+4}}}{\frac{1}{2\sqrt{x-1}}} = \frac{\sqrt{4}}{\sqrt{5}} = \frac{2}{\sqrt{5}} \] ### 13. \( \lim _{x \rightarrow \frac{\pi}{3}} \frac{\cos x-\frac{1}{2}}{x-\frac{\pi}{3}} \) Substituting \( x