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Math  /  Calculus

Question2. Evaluate the following lim⁡x→∞x310(10x−sin⁡10x)\lim _{x \rightarrow \infty} \frac{x^{3}}{10}\left(\frac{10}{x}-\sin \frac{10}{x}\right)

Studdy Solution

STEP 1

1. We are evaluating the limit as x x approaches infinity.
2. The expression involves a polynomial and trigonometric function.
3. We will use the properties of limits and asymptotic behavior of functions.

STEP 2

1. Simplify the expression inside the limit.
2. Evaluate the limit using known limit properties and behaviors.

STEP 3

First, simplify the expression inside the limit:
x310(10x−sin⁡10x) \frac{x^{3}}{10}\left(\frac{10}{x}-\sin \frac{10}{x}\right)
Distribute x310 \frac{x^3}{10} into the terms inside the parentheses:
=x310⋅10x−x310⋅sin⁡10x = \frac{x^3}{10} \cdot \frac{10}{x} - \frac{x^3}{10} \cdot \sin \frac{10}{x}
Simplify each term:
=x2−x310⋅sin⁡10x = x^2 - \frac{x^3}{10} \cdot \sin \frac{10}{x}

STEP 4

Evaluate the limit of each term separately as x→∞ x \to \infty .
First term:
lim⁡x→∞x2=∞ \lim_{x \to \infty} x^2 = \infty
Second term:
Consider the behavior of sin⁡10x \sin \frac{10}{x} as x→∞ x \to \infty . Since 10x→0 \frac{10}{x} \to 0 , we have:
sin⁡10x≈10x \sin \frac{10}{x} \approx \frac{10}{x}
Thus, the second term becomes:
lim⁡x→∞x310⋅10x=lim⁡x→∞x2=∞ \lim_{x \to \infty} \frac{x^3}{10} \cdot \frac{10}{x} = \lim_{x \to \infty} x^2 = \infty

STEP 5

Combine the limits of the two terms:
lim⁡x→∞(x2−x2)=lim⁡x→∞0=0 \lim_{x \to \infty} \left( x^2 - x^2 \right) = \lim_{x \to \infty} 0 = 0
The value of the limit is:
0 \boxed{0}

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