Math Problem Statement

lim(x -> 0+) (cosx - 2x)/(2xsinx+x^2cosx)

Solution

We are tasked with finding the limit:

limx0+cos(x)2x2xsin(x)+x2cos(x)\lim_{x \to 0^+} \frac{\cos(x) - 2x}{2x \sin(x) + x^2 \cos(x)}

Step 1: Direct Substitution

First, let's try direct substitution by plugging in x=0x = 0 into both the numerator and the denominator:

Numerator:

cos(0)2(0)=10=1\cos(0) - 2(0) = 1 - 0 = 1

Denominator:

2(0)sin(0)+(0)2cos(0)=0+0=02(0) \sin(0) + (0)^2 \cos(0) = 0 + 0 = 0

We have a form of 10\frac{1}{0}, which is undefined, suggesting that we may need to use L'Hopital's Rule, since we encounter a division by zero.

Step 2: Apply L'Hopital's Rule

L'Hopital's Rule is applicable for indeterminate forms like 00\frac{0}{0} or \frac{\infty}{\infty}. To apply it, we differentiate the numerator and the denominator separately with respect to xx.

Derivative of the Numerator:

The numerator is cos(x)2x\cos(x) - 2x, so the derivative is: ddx[cos(x)2x]=sin(x)2\frac{d}{dx}[\cos(x) - 2x] = -\sin(x) - 2

Derivative of the Denominator:

The denominator is 2xsin(x)+x2cos(x)2x \sin(x) + x^2 \cos(x). Using the product rule for both terms, the derivative is: ddx[2xsin(x)+x2cos(x)]=2sin(x)+2xcos(x)+2xcos(x)x2sin(x)\frac{d}{dx}[2x \sin(x) + x^2 \cos(x)] = 2 \sin(x) + 2x \cos(x) + 2x \cos(x) - x^2 \sin(x) =2sin(x)+4xcos(x)x2sin(x)= 2 \sin(x) + 4x \cos(x) - x^2 \sin(x)

Now, we have a new expression for the limit:

limx0+sin(x)22sin(x)+4xcos(x)x2sin(x)\lim_{x \to 0^+} \frac{-\sin(x) - 2}{2 \sin(x) + 4x \cos(x) - x^2 \sin(x)}

Step 3: Evaluate the New Limit

Now, we substitute x=0x = 0 into the new expressions.

Numerator:

sin(0)2=02=2-\sin(0) - 2 = 0 - 2 = -2

Denominator:

2sin(0)+4(0)cos(0)(0)2sin(0)=0+00=02 \sin(0) + 4(0) \cos(0) - (0)^2 \sin(0) = 0 + 0 - 0 = 0

We still get a form of 20\frac{-2}{0}, which suggests that the limit is infinite. Specifically, the sign of the denominator will tell us whether it approaches ++\infty or -\infty.

Step 4: Analyze the Denominator's Behavior

For small xx, we can approximate the terms in the denominator as follows:

  • sin(x)x\sin(x) \approx x (for small xx)
  • cos(x)1\cos(x) \approx 1 (for small xx)

So, for small xx, the denominator becomes approximately:

2x+4x0=6x2x + 4x - 0 = 6x

Thus, the denominator approaches 6x6x as x0+x \to 0^+, and the numerator is 2-2. Therefore, the limit is:

limx0+26x=\lim_{x \to 0^+} \frac{-2}{6x} = -\infty

Final Answer:

limx0+cos(x)2x2xsin(x)+x2cos(x)=\lim_{x \to 0^+} \frac{\cos(x) - 2x}{2x \sin(x) + x^2 \cos(x)} = -\infty

Would you like to go over any part of the solution in more detail, or is there another question I can help with?


Here are some related questions you might find interesting:

  1. How does L'Hopital's Rule work for limits of the form 00\frac{0}{0}?
  2. What happens when the denominator tends to 0 while the numerator tends to a non-zero value?
  3. Can we use Taylor series expansions to approximate the behavior of the functions involved?
  4. How does the approximation sin(x)x\sin(x) \approx x and cos(x)1\cos(x) \approx 1 help in evaluating limits?
  5. How do we handle limits involving trigonometric functions in general?

Tip: When dealing with limits that result in indeterminate forms, always check if L'Hopital's Rule can be applied or if approximations (like small angle approximations for sine and cosine) can simplify the process.

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Math Problem Analysis

Mathematical Concepts

Limits
Trigonometric Functions
L'Hopital's Rule
Approximation of Trigonometric Functions

Formulas

L'Hopital's Rule: lim(x -> c) f(x)/g(x) = lim(x -> c) f'(x)/g'(x) if the limit is indeterminate
Small Angle Approximation: sin(x) ≈ x, cos(x) ≈ 1

Theorems

L'Hopital's Rule

Suitable Grade Level

Grades 11-12