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Chapter 1: Q. 28 (page 153)

Find the limit by hand.
$\lim_{x \to 0} \frac{x^{2}}{e^{x} - 1}$

Short Answer

Expert verified

On solving the limit, we get, $0$ .

Step by step solution

01

Step 1. Given information.

Consider the given question,

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

02

Step 2. Solve the given limit.

Rewriting the expression,

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

Using the product rule,

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

As $\lim_{h \to 0} \frac{h}{e^{h} - 1} = 1$ ,

$= 1 (0) = 0$

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Most popular questions from this chapter

Write delta-epsilon proofs for each of the limit statements $\lim_{x \to c} f (x) = L$ in Exercises $47 - 60 .$

$\lim_{x \to 1} \frac{x^{2} - 1}{x - 1} = 2$

Calculate each of the limits:

$\lim_{x \to 1} x \sin^{- 1} \frac{x}{2}$ .

For each limit statement , use algebra to find δ > 0 in terms of $ε$ > 0 so that if 0 < |x − c| < δ, then | f(x) − L| < $ε$ .

$\lim_{x \to 3} (x^{2} - 6 x + 5) = - 4$

What are punctured intervals, and why do we need to use them when discussing limits?

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

$\lim_{x \to - 5} (3 x - 2) .$

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