Question 8 of 14intermediate🔧 ApplyNumerical2 marks

Evaluate the following limit, if exist: lim⁡x→0e2+x−e2x\lim_{x \to 0} \frac{e^{2+x} - e^2}{x}

Correct Answer

e2e^2

Exercise: Exercise 13.2 | Q: 2 | (Page Page 6)
For More Understanding

Explanation

This question from Exercise 13.2 tests the application of the standard limit result lim⁡x→0ex−1x=1\lim_{x \to 0} \frac{e^x - 1}{x} = 1 (Theorem 6). The key insight is to factor out e2e^2 from the numerator, which transforms the expression into a form where the standard limit can be directly applied.

Solution Steps

  1. Step 1: Factor out e2e^2 from the numerator: lim⁡x→0e2+x−e2x=lim⁡x→0e2⋅ex−e2x=lim⁡x→0e2(ex−1)x\lim_{x \to 0} \frac{e^{2+x} - e^2}{x} = \lim_{x \to 0} \frac{e^2 \cdot e^x - e^2}{x} = \lim_{x \to 0} \frac{e^2(e^x - 1)}{x}

  2. Step 2: Separate the constant from the limit: =e2⋅lim⁡x→0ex−1x= e^2 \cdot \lim_{x \to 0} \frac{e^x - 1}{x}

  3. Step 3: Apply the standard result lim⁡x→0ex−1x=1\lim_{x \to 0} \frac{e^x - 1}{x} = 1 (Theorem 6): =e2⋅1=e2= e^2 \cdot 1 = e^2

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