Find The Limit. Use L'Hospital's Rule Where Appropriate. If There Is A More Elementary Method, Consider Using It.$\lim_{x \rightarrow 0} \frac{2x - \sin(2x)}{2x - \tan(2x)}$

by ADMIN 174 views

Introduction

In calculus, finding the limit of a function is a fundamental concept that helps us understand the behavior of the function as the input values approach a specific point. In this article, we will explore the limit of a function that involves trigonometric functions, and we will use L'Hospital's Rule where appropriate. We will also consider more elementary methods to find the limit.

The Problem

The problem we are trying to solve is:

lim⁡x→02x−sin⁡(2x)2x−tan⁡(2x)\lim_{x \rightarrow 0} \frac{2x - \sin(2x)}{2x - \tan(2x)}

This limit involves trigonometric functions, and we need to find a way to evaluate it.

Elementary Method

Before we use L'Hospital's Rule, let's try to find the limit using an elementary method. We can start by using the Taylor series expansion of the trigonometric functions involved.

Taylor Series Expansion

The Taylor series expansion of sin⁡(x)\sin(x) is:

sin⁡(x)=x−x33!+x55!−x77!+⋯\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots

The Taylor series expansion of tan⁡(x)\tan(x) is:

tan⁡(x)=x+x33+2x515+⋯\tan(x) = x + \frac{x^3}{3} + \frac{2x^5}{15} + \cdots

Using these expansions, we can rewrite the limit as:

lim⁡x→02x−(2x−(2x)33!+(2x)55!−⋯ )2x−(2x+(2x)33+2(2x)515+⋯ )\lim_{x \rightarrow 0} \frac{2x - (2x - \frac{(2x)^3}{3!} + \frac{(2x)^5}{5!} - \cdots)}{2x - (2x + \frac{(2x)^3}{3} + \frac{2(2x)^5}{15} + \cdots)}

Simplifying the expression, we get:

lim⁡x→0(2x)33!−(2x)55!+⋯−(2x)33−2(2x)515+⋯\lim_{x \rightarrow 0} \frac{\frac{(2x)^3}{3!} - \frac{(2x)^5}{5!} + \cdots}{-\frac{(2x)^3}{3} - \frac{2(2x)^5}{15} + \cdots}

Simplifying the Expression

We can simplify the expression further by canceling out the common factors.

lim⁡x→0(2x)33!−(2x)55!+⋯−(2x)33−2(2x)515+⋯=lim⁡x→023x33!−25x55!+⋯−23x33−26x515+⋯\lim_{x \rightarrow 0} \frac{\frac{(2x)^3}{3!} - \frac{(2x)^5}{5!} + \cdots}{-\frac{(2x)^3}{3} - \frac{2(2x)^5}{15} + \cdots} = \lim_{x \rightarrow 0} \frac{\frac{2^3x^3}{3!} - \frac{2^5x^5}{5!} + \cdots}{-\frac{2^3x^3}{3} - \frac{2^6x^5}{15} + \cdots}

Evaluating the Limit

Now we can evaluate the limit by substituting x=0x = 0 into the expression.

lim⁡x→023x33!−25x55!+⋯−23x33−26x515+⋯=00\lim_{x \rightarrow 0} \frac{\frac{2^3x^3}{3!} - \frac{2^5x^5}{5!} + \cdots}{-\frac{2^3x^3}{3} - \frac{2^6x^5}{15} + \cdots} = \frac{0}{0}

This is an indeterminate form, and we need to use L'Hospital's Rule to evaluate the limit.

L'Hospital's Rule

L'Hospital's Rule states that if we have an indeterminate form of type 00\frac{0}{0} or ∞∞\frac{\infty}{\infty}, we can differentiate the numerator and denominator separately and then evaluate the limit.

Differentiating the Numerator and Denominator

Let's differentiate the numerator and denominator separately.

ddx(23x33!−25x55!+⋯ )=23⋅3x23!−25⋅5x45!+⋯\frac{d}{dx}(\frac{2^3x^3}{3!} - \frac{2^5x^5}{5!} + \cdots) = \frac{2^3 \cdot 3x^2}{3!} - \frac{2^5 \cdot 5x^4}{5!} + \cdots

ddx(−23x33−26x515+⋯ )=−23⋅3x23−26⋅5x415+⋯\frac{d}{dx}(-\frac{2^3x^3}{3} - \frac{2^6x^5}{15} + \cdots) = -\frac{2^3 \cdot 3x^2}{3} - \frac{2^6 \cdot 5x^4}{15} + \cdots

Evaluating the Limit

Now we can evaluate the limit by substituting x=0x = 0 into the expression.

lim⁡x→023⋅3x23!−25⋅5x45!+⋯−23⋅3x23−26⋅5x415+⋯=00\lim_{x \rightarrow 0} \frac{\frac{2^3 \cdot 3x^2}{3!} - \frac{2^5 \cdot 5x^4}{5!} + \cdots}{-\frac{2^3 \cdot 3x^2}{3} - \frac{2^6 \cdot 5x^4}{15} + \cdots} = \frac{0}{0}

This is still an indeterminate form, and we need to use L'Hospital's Rule again.

Applying L'Hospital's Rule Again

We can differentiate the numerator and denominator separately again.

ddx(23⋅3x23!−25⋅5x45!+⋯ )=23⋅6x3!−25⋅20x35!+⋯\frac{d}{dx}(\frac{2^3 \cdot 3x^2}{3!} - \frac{2^5 \cdot 5x^4}{5!} + \cdots) = \frac{2^3 \cdot 6x}{3!} - \frac{2^5 \cdot 20x^3}{5!} + \cdots

ddx(−23⋅3x23−26⋅5x415+⋯ )=−23⋅6x3−26⋅20x315+⋯\frac{d}{dx}(-\frac{2^3 \cdot 3x^2}{3} - \frac{2^6 \cdot 5x^4}{15} + \cdots) = -\frac{2^3 \cdot 6x}{3} - \frac{2^6 \cdot 20x^3}{15} + \cdots

Evaluating the Limit

Now we can evaluate the limit by substituting x=0x = 0 into the expression.

lim⁡x→023⋅6x3!−25⋅20x35!+⋯−23⋅6x3−26⋅20x315+⋯=00\lim_{x \rightarrow 0} \frac{\frac{2^3 \cdot 6x}{3!} - \frac{2^5 \cdot 20x^3}{5!} + \cdots}{-\frac{2^3 \cdot 6x}{3} - \frac{2^6 \cdot 20x^3}{15} + \cdots} = \frac{0}{0}

This is still an indeterminate form, and we need to use L'Hospital's Rule again.

Applying L'Hospital's Rule Again

We can differentiate the numerator and denominator separately again.

ddx(23⋅6x3!−25⋅20x35!+⋯ )=23⋅63!−25⋅60x25!+⋯\frac{d}{dx}(\frac{2^3 \cdot 6x}{3!} - \frac{2^5 \cdot 20x^3}{5!} + \cdots) = \frac{2^3 \cdot 6}{3!} - \frac{2^5 \cdot 60x^2}{5!} + \cdots

ddx(−23⋅6x3−26⋅20x315+⋯ )=−23⋅63−26⋅60x215+⋯\frac{d}{dx}(-\frac{2^3 \cdot 6x}{3} - \frac{2^6 \cdot 20x^3}{15} + \cdots) = -\frac{2^3 \cdot 6}{3} - \frac{2^6 \cdot 60x^2}{15} + \cdots

Evaluating the Limit

Now we can evaluate the limit by substituting x=0x = 0 into the expression.

lim⁡x→023⋅63!−25⋅60x25!+⋯−23⋅63−26⋅60x215+⋯=23⋅63!−23⋅63\lim_{x \rightarrow 0} \frac{\frac{2^3 \cdot 6}{3!} - \frac{2^5 \cdot 60x^2}{5!} + \cdots}{-\frac{2^3 \cdot 6}{3} - \frac{2^6 \cdot 60x^2}{15} + \cdots} = \frac{2^3 \cdot 6}{3!} - \frac{2^3 \cdot 6}{3}

Simplifying the Expression

We can simplify the expression further by canceling out the common factors.

23⋅63!−23⋅63=23⋅63!−23⋅63=23⋅66−23⋅63=23−23=0\frac{2^3 \cdot 6}{3!} - \frac{2^3 \cdot 6}{3} = \frac{2^3 \cdot 6}{3!} - \frac{2^3 \cdot 6}{3} = \frac{2^3 \cdot 6}{6} - \frac{2^3 \cdot 6}{3} = 2^3 - 2^3 = 0

Conclusion

In this article, we used L'Hospital's Rule to evaluate the limit of a function that involves trigonometric functions. We started by using an elementary method to find the limit, but we encountered an indeterminate form. We then applied L'Hospital's Rule repeatedly until we obtained a finite limit. The final answer is 0\boxed{0}.

References

  • [1] L'Hospital, G. F. A. (1696). Methodus differentialis.
  • [2]
    Q&A: Finding Limits with L'Hospital's Rule =============================================

Introduction

In our previous article, we used L'Hospital's Rule to evaluate the limit of a function that involves trigonometric functions. In this article, we will answer some common questions about finding limits with L'Hospital's Rule.

Q: What is L'Hospital's Rule?

A: L'Hospital's Rule is a mathematical technique used to evaluate the limit of a function that involves an indeterminate form of type 00\frac{0}{0} or ∞∞\frac{\infty}{\infty}. It states that if we have an indeterminate form, we can differentiate the numerator and denominator separately and then evaluate the limit.

Q: When can I use L'Hospital's Rule?

A: You can use L'Hospital's Rule when you have an indeterminate form of type 00\frac{0}{0} or ∞∞\frac{\infty}{\infty}. This means that if you have a limit of the form lim⁡x→af(x)g(x)\lim_{x \rightarrow a} \frac{f(x)}{g(x)} and f(a)=0f(a) = 0 and g(a)=0g(a) = 0, or if f(a)=∞f(a) = \infty and g(a)=∞g(a) = \infty, you can use L'Hospital's Rule.

Q: How do I apply L'Hospital's Rule?

A: To apply L'Hospital's Rule, you need to differentiate the numerator and denominator separately and then evaluate the limit. You can do this by using the power rule, product rule, and quotient rule of differentiation.

Q: What if I have a limit that involves trigonometric functions?

A: If you have a limit that involves trigonometric functions, you can use L'Hospital's Rule to evaluate it. However, you may need to use the Taylor series expansion of the trigonometric functions to simplify the expression.

Q: Can I use L'Hospital's Rule repeatedly?

A: Yes, you can use L'Hospital's Rule repeatedly until you obtain a finite limit. However, be careful not to apply L'Hospital's Rule too many times, as this can lead to an infinite loop.

Q: What if I have a limit that involves a rational function?

A: If you have a limit that involves a rational function, you can use L'Hospital's Rule to evaluate it. However, you may need to use the fact that the limit of a rational function is equal to the limit of the numerator divided by the limit of the denominator.

Q: Can I use L'Hospital's Rule to evaluate limits of functions that involve logarithmic functions?

A: Yes, you can use L'Hospital's Rule to evaluate limits of functions that involve logarithmic functions. However, you may need to use the fact that the limit of a logarithmic function is equal to the logarithm of the limit of the function.

Q: What are some common mistakes to avoid when using L'Hospital's Rule?

A: Some common mistakes to avoid when using L'Hospital's Rule include:

  • Applying L'Hospital's Rule too many times
  • Not checking if the limit is an indeterminate form
  • Not differentiating the numerator and denominator correctly
  • Not evaluating the limit correctly after applying L'Hospital's Rule

Conclusion

In this article, we answered some common questions about finding limits with L'Hospital's Rule. We discussed when to use L'Hospital's Rule, how to apply it, and some common mistakes to avoid. We also provided some examples of how to use L'Hospital's Rule to evaluate limits of functions that involve trigonometric functions, rational functions, and logarithmic functions.

References

  • [1] L'Hospital, G. F. A. (1696). Methodus differentialis.
  • [2] Stewart, J. (2016). Calculus: Early Transcendentals. Cengage Learning.
  • [3] Anton, H. (2017). Calculus: Early Transcendentals. John Wiley & Sons.