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Chapter 1: Q. 51 (page 136)

$\lim_{x \to 0} (\frac{e^{x} - 1}{e^{2 x} + 2 e^{x} - 3})$

Short Answer

Expert verified

$\lim_{x \to 0} (\frac{e^{x} - 1}{e^{2 x} + 2 e^{x} - 3}) = \frac{1}{4}$

Step by step solution

01

Step 1. Given information

$\lim_{x \to 0} (\frac{e^{x} - 1}{e^{2 x} + 2 e^{x} - 3})$

02

Step 2. Plug in the values

$\lim_{x \to 0} (\frac{e^{x} - 1}{e^{2 x} + 2 e^{x} - 3}) = \lim_{x \to 0} (\frac{1}{e^{x} + 3})$

$\lim_{x \to 0} (\frac{e^{x} - 1}{e^{2 x} + 2 e^{x} - 3}) = \frac{1}{e^{0} + 3}$

$\lim_{x \to 0} (\frac{e^{x} - 1}{e^{2 x} + 2 e^{x} - 3}) = \frac{1}{4}$

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Most popular questions from this chapter

Write a delta–epsilon proof that proves that $f$ is continuous on its domain. In each case, you will need to assume that δ is less than or equal to $1$ .

$f (x) = x^{2}$

Calculate each of the limits:

$\lim_{h \to 0} \frac{{(- 1 + h)}^{3} - {(- 1)}^{3}}{h}$ .

For each limit in Exercises 33–38, either use continuity to calculate the limit or explain why Theorem 1.16 does not apply.

$\lim_{x \to - 1} 6 .$

State what it means for a function f to be right continuous at a point x = c, in terms of the delta–epsilon definition of limit.

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

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