Question 14 of 14intermediate🔧 ApplyNumerical3 marks

Evaluate the following limit, if exist: lim⁡x→0log⁡(1+x3)sin⁡3x\lim_{x \to 0} \frac{\log (1 + x^3)}{\sin^3 x}

Correct Answer

1

Exercise: Exercise 13.2 | Q: 8 | (Page Page 6)
For More Understanding

Explanation

This limit evaluates to 1 by applying two standard limits from the textbook. The key insight is to rewrite the expression to separate the logarithmic and trigonometric parts, then apply the standard limits lim⁡x→0log⁡(1+x)x=1\lim_{x \to 0} \frac{\log(1+x)}{x} = 1 (Theorem 7) and lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1 (from Example 6).

Solution Steps

  1. Step 1: Rewrite the expression by multiplying and dividing by x3x^3:

    lim⁡x→0log⁡(1+x3)sin⁡3x=lim⁡x→0log⁡(1+x3)x3⋅x3sin⁡3x\lim_{x \to 0} \frac{\log (1 + x^3)}{\sin^3 x} = \lim_{x \to 0} \frac{\log (1 + x^3)}{x^3} \cdot \frac{x^3}{\sin^3 x}

  2. Step 2: Evaluate the first part. As x→0x \to 0, x3→0x^3 \to 0. Using the standard limit lim⁡y→0log⁡(1+y)y=1\lim_{y \to 0} \frac{\log(1+y)}{y} = 1 with y=x3y = x^3:

    lim⁡x→0log⁡(1+x3)x3=1\lim_{x \to 0} \frac{\log (1 + x^3)}{x^3} = 1

  3. Step 3: Evaluate the second part. Using lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1:

    lim⁡x→0x3sin⁡3x=lim⁡x→0(xsin⁡x)3=13=1\lim_{x \to 0} \frac{x^3}{\sin^3 x} = \lim_{x \to 0} \left(\frac{x}{\sin x}\right)^3 = 1^3 = 1

  4. Step 4: Multiply both results:

    lim⁡x→0log⁡(1+x3)sin⁡3x=1×1=1\lim_{x \to 0} \frac{\log (1 + x^3)}{\sin^3 x} = 1 \times 1 = 1
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