Math Problem Statement

Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it. lim x→0+ (cos(x))3/x2

Solution

To find the limit

limx0+(cos(x))3x2,\lim_{x \to 0^+} \frac{(\cos(x))^3}{x^2},

we first substitute x=0x = 0:

(cos(0))302=130=10,\frac{(\cos(0))^3}{0^2} = \frac{1^3}{0} = \frac{1}{0},

which indicates that the limit approaches infinity. However, to confirm whether the limit diverges or has a specific behavior, we can analyze the function further.

As xx approaches 0+0^+, cos(x)\cos(x) approaches 11. Therefore, (cos(x))3(\cos(x))^3 approaches 11 as well. The denominator x2x^2 approaches 00. Thus, we have:

(cos(x))3x210+=+.\frac{(\cos(x))^3}{x^2} \to \frac{1}{0^+} = +\infty.

To use l'Hôpital's Rule, we can re-examine the limit. Since both the numerator and denominator approach 00 as x0+x \to 0^+, we can rewrite it as follows:

  1. Apply l'Hôpital's Rule:

We differentiate the numerator and denominator:

Numerator: ddx((cos(x))3)=3(cos(x))2(sin(x))=3(cos(x))2sin(x),\text{Numerator: } \frac{d}{dx} \left((\cos(x))^3\right) = 3(\cos(x))^2 (-\sin(x)) = -3(\cos(x))^2 \sin(x),

Denominator: ddx(x2)=2x.\text{Denominator: } \frac{d}{dx}(x^2) = 2x.

Thus, we have:

limx0+(cos(x))3x2=limx0+3(cos(x))2sin(x)2x.\lim_{x \to 0^+} \frac{(\cos(x))^3}{x^2} = \lim_{x \to 0^+} \frac{-3(\cos(x))^2 \sin(x)}{2x}.

Now substituting x=0x = 0:

The numerator approaches 00 (since sin(0)=0\sin(0) = 0) and the denominator approaches 00 as well, allowing us to apply l'Hôpital's Rule again.

  1. Apply l'Hôpital's Rule a second time:

Differentiate again:

  • The numerator becomes:

ddx(3(cos(x))2sin(x))=3(2cos(x)(sin(x))sin(x)+(cos(x))2cos(x))=3(2cos(x)sin2(x)cos3(x)),\frac{d}{dx} \left(-3(\cos(x))^2 \sin(x)\right) = -3 \left(2\cos(x)(-\sin(x)) \sin(x) + (\cos(x))^2 \cos(x)\right) = 3 \left(2\cos(x)\sin^2(x) - \cos^3(x)\right),

  • The denominator becomes:

ddx(2x)=2.\frac{d}{dx}(2x) = 2.

Now the limit is:

limx0+3(2cos(x)sin2(x)cos3(x))2.\lim_{x \to 0^+} \frac{3(2\cos(x)\sin^2(x) - \cos^3(x))}{2}.

As x0+x \to 0^+, this evaluates to:

limx0+3(2101)2=3(1)2=32.\lim_{x \to 0^+} \frac{3(2 \cdot 1 \cdot 0 - 1)}{2} = \frac{3(-1)}{2} = -\frac{3}{2}.

However, since we want the limit of the original function as x0+x \to 0^+, we find:

limx0+(cos(x))3x2=+.\lim_{x \to 0^+} \frac{(\cos(x))^3}{x^2} = +\infty.

Thus, the final answer is:

limx0+(cos(x))3x2=+.\lim_{x \to 0^+} \frac{(\cos(x))^3}{x^2} = +\infty.

Summary:

The limit diverges to positive infinity.

Do you want further details on this solution or have any questions? Here are some related questions you might consider:

  1. What is the behavior of cos(x)\cos(x) as xx approaches other values?
  2. How does l'Hôpital's Rule apply to different types of indeterminate forms?
  3. Can you explain the Taylor series expansion for cos(x)\cos(x)?
  4. What are other limits involving trigonometric functions?
  5. How does the limit change if we consider limx0(cos(x))3x2\lim_{x \to 0} \frac{(\cos(x))^3}{x^2} from both sides?

Tip: Always check the form of the limit before deciding on the method to apply!

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Math Problem Analysis

Mathematical Concepts

Limits
Trigonometric Functions
L'Hôpital's Rule

Formulas

lim (cos(x))^3/x^2 as x→0+
L'Hôpital's Rule: lim f(x)/g(x) = lim f'(x)/g'(x) if indeterminate

Theorems

L'Hôpital's Rule

Suitable Grade Level

Grades 11-12