Question 11 of 14intermediate🔧 ApplyNumerical2 marks

Evaluate the following limit, if exist: lim⁡x→3ex−e3x−3\lim_{x \to 3} \frac{e^x - e^3}{x - 3}

Correct Answer

e3e^3

Exercise: Exercise 13.2 | Q: 5 | (Page Page 6)
For More Understanding

Explanation

The limit is evaluated by substituting x=3+hx = 3 + h, factoring e3e^3, and applying the standard limit lim⁡h→0eh−1h=1\lim_{h \to 0} \frac{e^h - 1}{h} = 1 (Theorem 6 in the provided context). This matches the answer given for Exercise 13.2, question 5.

Solution Steps

  1. Step 1: Let x=3+hx = 3 + h. Then as x→3x \to 3, h→0h \to 0. The limit becomes lim⁡h→0e3+h−e3h\lim_{h \to 0} \frac{e^{3+h} - e^3}{h}.

  2. Step 2: Factor out e3e^3: =lim⁡h→0e3(eh−1)h=e3lim⁡h→0eh−1h= \lim_{h \to 0} \frac{e^3(e^h - 1)}{h} = e^3 \lim_{h \to 0} \frac{e^h - 1}{h}.

  3. Step 3: Using the standard limit lim⁡h→0eh−1h=1\lim_{h \to 0} \frac{e^h - 1}{h} = 1 (Theorem 6), we obtain e3⋅1=e3e^3 \cdot 1 = e^3.

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