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Chapter 1: Q. 53 (page 136)

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

Short Answer

Expert verified

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

Step by step solution

01

Step 1. Given information

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

02

Step 2.Simplify

Since,

\lim_{x \to a} [c \cdot f (x)] = c \cdot \lim_{x \to a} f (x)

Therefore,

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

03

Step 3. Calculate the value

Apply L'Hopital's rule,

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

role="math" localid="1648011355231" $\lim_{x \to 0} (\frac{2 x}{e^{x} - 1}) = 2 \frac{1}{e^{0}}$

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

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Most popular questions from this chapter

Write each of the inequalities in interval notation:

$0 < | x - 2 | < 0.1$

In Exercises 39–44, use Theorem 1.16 and left and right limits to determine whether each function f is continuous at its break point(s). For each discontinuity of f, describe the type of discontinuity and any one-sided discontinuity.

$f (x) = \{\begin{matrix} x + 1, i f x < 1 \\ 3 x - 1, i f 1 ⩽ x < 2 \\ x + 2, i f x ⩾ 2 \end{matrix} .$

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

$\lim_{x \to 3} x^{4} .$

If $f$ is a continuous function, what can you say about $\lim_{x \to 1} f (x) ?$

For each limit statement in Exercises $41 - 44$ , use algebra to find $δ$ or $N$ in terms of $ε$ or $M$ , according to the appropriate formal limit definition.

$\lim_{x \to \infty} (x^{2} + 2) = \infty$ ,find $N$ in terms of $M$ .

See all solutions

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