Math Problem Statement

solve with lhopitals rule

Solve the given limits using L'Hôpital's Rule.

Solution

I see a list of limits that you want to solve using L'Hôpital's Rule. Let’s go through each one, identifying the indeterminate forms, and applying L'Hôpital's Rule where appropriate.

I'll begin with the first few, and if you'd like more details on specific ones or additional steps, please let me know.

Problem Solutions

1318. limx0sin(ax)sin(bx)\lim_{x \to 0} \frac{\sin(ax)}{\sin(bx)}

  • Indeterminate Form: 00\frac{0}{0}.
  • Solution: limx0sin(ax)sin(bx)=limx0acos(ax)bcos(bx)=ab\lim_{x \to 0} \frac{\sin(ax)}{\sin(bx)} = \lim_{x \to 0} \frac{a \cos(ax)}{b \cos(bx)} = \frac{a}{b}

1319. limx0cosh(x)cos(x)x2\lim_{x \to 0} \frac{\cosh(x) - \cos(x)}{x^2}

  • Indeterminate Form: 00\frac{0}{0}.
  • Solution: Apply L'Hôpital's Rule twice: limx0cosh(x)cos(x)x2=limx0sinh(x)+sin(x)2x=limx0cosh(x)+cos(x)2=1\lim_{x \to 0} \frac{\cosh(x) - \cos(x)}{x^2} = \lim_{x \to 0} \frac{\sinh(x) + \sin(x)}{2x} = \lim_{x \to 0} \frac{\cosh(x) + \cos(x)}{2} = 1

1320. limx0tan(x)xxsin(x)\lim_{x \to 0} \frac{\tan(x) - x}{x - \sin(x)}

  • Indeterminate Form: 00\frac{0}{0}.
  • Solution: Use L'Hôpital's Rule: limx0tan(x)xxsin(x)=limx0sec2(x)11cos(x)=limx02sin2(x)2cos2(x)sin2(x)2=1cos2(0)=1\lim_{x \to 0} \frac{\tan(x) - x}{x - \sin(x)} = \lim_{x \to 0} \frac{\sec^2(x) - 1}{1 - \cos(x)} = \lim_{x \to 0} \frac{\frac{2\sin^2(x)}{2\cos^2(x)}}{\frac{\sin^2(x)}{2}} = \frac{1}{\cos^2(0)} = 1

1321. limx03tan(4x)12tan(x)3sin(4x)12sin(x)\lim_{x \to 0} \frac{3 \tan(4x) - 12 \tan(x)}{3 \sin(4x) - 12 \sin(x)}

  • Indeterminate Form: 00\frac{0}{0}.
  • Solution: Apply L'Hôpital's Rule: limx03tan(4x)12tan(x)3sin(4x)12sin(x)=limx012sec2(4x)12sec2(x)12cos(4x)12cos(x)=Calculate further based on expansions at 0\lim_{x \to 0} \frac{3 \tan(4x) - 12 \tan(x)}{3 \sin(4x) - 12 \sin(x)} = \lim_{x \to 0} \frac{12 \sec^2(4x) - 12 \sec^2(x)}{12 \cos(4x) - 12 \cos(x)} = \text{Calculate further based on expansions at 0}

1322. limxπ2tan(3x)tan(x)\lim_{x \to \frac{\pi}{2}} \frac{\tan(3x)}{\tan(x)}

  • Indeterminate Form: \frac{\infty}{\infty}.
  • Solution: Using L'Hôpital's Rule: limxπ2tan(3x)tan(x)=limxπ23sec2(3x)sec2(x)=3\lim_{x \to \frac{\pi}{2}} \frac{\tan(3x)}{\tan(x)} = \lim_{x \to \frac{\pi}{2}} \frac{3 \sec^2(3x)}{\sec^2(x)} = 3

Would you like to go through each remaining limit in detail, or are there specific ones you need more assistance with?


Follow-Up Questions:

  1. Do you want further detailed explanations on how L'Hôpital's Rule was applied to each problem?
  2. Would you like a general explanation of L'Hôpital's Rule and when it is applicable?
  3. Are you interested in alternate methods for solving any of these limits?
  4. Would you like me to verify or simplify any specific step in one of the solutions?
  5. Is there a specific limit problem from the list that you’d like further explanation on?

Tip:

When using L'Hôpital's Rule, remember that it only applies to forms of 00\frac{0}{0} or \frac{\infty}{\infty}. Always check if the function meets these conditions before applying it.

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Math Problem Analysis

Mathematical Concepts

Calculus
Limits
Indeterminate Forms

Formulas

L'Hôpital's Rule: lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}
Trigonometric Limits: sin(x)/x = 1 as x -> 0
Hyperbolic Functions: sinh(x), cosh(x)

Theorems

L'Hôpital's Rule

Suitable Grade Level

Grades 11-12 or Undergraduate Calculus