Math Problem Statement

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

Solution

To evaluate the limit:

limx0+cosx2x2xsinx+x2cosx,\lim_{x \to 0^+} \frac{\cos x - 2x}{2x \sin x + x^2 \cos x},

we will proceed step-by-step.


Step 1: Substitution at x=0x = 0

Direct substitution of x=0x = 0 gives: cos(0)2(0)2(0)sin(0)+(0)2cos(0)=100+0=10.\frac{\cos(0) - 2(0)}{2(0) \sin(0) + (0)^2 \cos(0)} = \frac{1 - 0}{0 + 0} = \frac{1}{0}. This is indeterminate since the denominator approaches 00. We proceed with L’Hôpital’s Rule as both the numerator and denominator tend to 00 as x0+x \to 0^+.


Step 2: Apply L’Hôpital’s Rule

Differentiate the numerator and the denominator with respect to xx:

Numerator:

ddx(cosx2x)=sinx2.\frac{d}{dx}(\cos x - 2x) = -\sin x - 2.

Denominator:

ddx(2xsinx+x2cosx)=2sinx+2xcosx+2xcosxx2sinx.\frac{d}{dx}(2x \sin x + x^2 \cos x) = 2 \sin x + 2x \cos x + 2x \cos x - x^2 \sin x. Simplify the derivative of the denominator: ddx(2xsinx+x2cosx)=2sinx+4xcosxx2sinx.\frac{d}{dx}(2x \sin x + x^2 \cos x) = 2 \sin x + 4x \cos x - x^2 \sin x.


Step 3: Rewrite the limit

The limit becomes: limx0+sinx22sinx+4xcosxx2sinx.\lim_{x \to 0^+} \frac{-\sin x - 2}{2 \sin x + 4x \cos x - x^2 \sin x}.


Step 4: Substitution at x=0x = 0 again

Substituting x=0x = 0:

  • Numerator: sin(0)2=2-\sin(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.

Again, this is indeterminate, so we apply L’Hôpital’s Rule once more.


Step 5: Apply L’Hôpital’s Rule again

Differentiate the numerator and denominator again:

Numerator:

ddx(sinx2)=cosx.\frac{d}{dx}(-\sin x - 2) = -\cos x.

Denominator:

Differentiate 2sinx+4xcosxx2sinx2 \sin x + 4x \cos x - x^2 \sin x: ddx(2sinx+4xcosxx2sinx)=2cosx+4cosx4xsinx2xsinxx2cosx.\frac{d}{dx}(2 \sin x + 4x \cos x - x^2 \sin x) = 2 \cos x + 4 \cos x - 4x \sin x - 2x \sin x - x^2 \cos x. Simplify: ddx(2sinx+4xcosxx2sinx)=6cosx6xsinxx2cosx.\frac{d}{dx}(2 \sin x + 4x \cos x - x^2 \sin x) = 6 \cos x - 6x \sin x - x^2 \cos x.


Step 6: Rewrite the limit

The limit becomes: limx0+cosx6cosx6xsinxx2cosx.\lim_{x \to 0^+} \frac{-\cos x}{6 \cos x - 6x \sin x - x^2 \cos x}.


Step 7: Substitution at x=0x = 0

Substitute x=0x = 0:

  • Numerator: cos(0)=1-\cos(0) = -1,
  • Denominator: 6cos(0)6(0)sin(0)(0)2cos(0)=600=66 \cos(0) - 6(0) \sin(0) - (0)^2 \cos(0) = 6 - 0 - 0 = 6.

Thus, the limit is: limx0+cosx6cosx6xsinxx2cosx=16.\lim_{x \to 0^+} \frac{-\cos x}{6 \cos x - 6x \sin x - x^2 \cos x} = \frac{-1}{6}.


Final Answer:

16\boxed{-\frac{1}{6}}

Let me know if you would like further clarification or additional examples. 😊


Five Related Questions:

  1. How is L’Hôpital’s Rule applied to solve indeterminate forms?
  2. What are common indeterminate forms encountered in calculus?
  3. Can you evaluate limx0+sinxx2+cosx1\lim_{x \to 0^+} \frac{\sin x}{x^2 + \cos x - 1}?
  4. How do trigonometric approximations (e.g., sinxx\sin x \approx x for small xx) help in solving limits?
  5. What are alternative methods to solve limits when L’Hôpital’s Rule fails?

Tip:

When using L’Hôpital’s Rule, always check that the function is in an indeterminate form like 00\frac{0}{0} or \frac{\infty}{\infty} before applying the rule.

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Math Problem Analysis

Mathematical Concepts

Limits
L'Hopital's Rule
Trigonometric Functions
Indeterminate Forms

Formulas

L'Hopital's Rule
cos(x)
sin(x)

Theorems

L'Hopital's Rule

Suitable Grade Level

Grades 11-12