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Chapter 1: Q. 52 (page 136)

$\lim_{x \to 1} (\frac{e^{x} - 1}{e^{2 x} + 2 e^{x} - 3})$

Short Answer

Expert verified

$\lim_{x \to 1} (\frac{e^{x} - 1}{e^{2 x} + 2 e^{x} - 3}) = \frac{e - 1}{e^{2} + 2 e - 3}$

Step by step solution

Step 1. Given information

$\lim_{x \to 1} (\frac{e^{x} - 1}{e^{2 x} + 2 e^{x} - 3})$

Step 2. Plug in the values

$\lim_{x \to 1} (\frac{e^{x} - 1}{e^{2 x} + 2 e^{x} - 3}) = \frac{e^{1} - 1}{e^{2 \cdot 1} + 2 e^{1} - 3}$

$\lim_{x \to 1} (\frac{e^{x} - 1}{e^{2 x} + 2 e^{x} - 3}) = \frac{e - 1}{e^{2} + 2 e - 3}$

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limx→1ex-1e2x+2ex-3

Short Answer

Step by step solution

Step 1. Given information

Step 2. Plug in the values

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$\lim_{x \to 1} (\frac{e^{x} - 1}{e^{2 x} + 2 e^{x} - 3})$