Question 10 of 14intermediate🔧 ApplyNumerical2 marks

Evaluate the following limit, if exist: lim⁡x→0esin⁡x−1x\lim_{x \to 0} \frac{e^{\sin x} - 1}{x}

Correct Answer

1

Exercise: Exercise 13.2 | Q: 4 | (Page Page 6)
For More Understanding

Explanation

This question from Exercise 13.2 (Question 4) tests the application of standard limits. The solution uses two fundamental results: Theorem 6 establishes that lim(x→0) (exe^{x} - 1)/x = 1, and Example 6 uses the standard limit lim(x→0) sin x/x = 1. By multiplying and dividing by sin x, the expression splits into two parts, each evaluating to 1.

Solution Steps

  1. Step 1: Multiply and divide by sin x:

    lim⁡x→0esin⁡x−1x=lim⁡x→0esin⁡x−1sin⁡x×sin⁡xx\lim_{x \to 0} \frac{e^{\sin x} - 1}{x} = \lim_{x \to 0} \frac{e^{\sin x} - 1}{\sin x} \times \frac{\sin x}{x}

  2. Step 2: As x → 0, sin x → 0. Using the standard limit lim⁡y→0ey−1y=1\lim_{y \to 0} \frac{e^y - 1}{y} = 1:

    lim⁡x→0esin⁡x−1sin⁡x=1\lim_{x \to 0} \frac{e^{\sin x} - 1}{\sin x} = 1

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

    lim⁡x→0esin⁡x−1x=1×1=1\lim_{x \to 0} \frac{e^{\sin x} - 1}{x} = 1 \times 1 = 1
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