Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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Could you do 15 and 17, thanks

**Chapter 2: Limits of Functions**

**15.** Let \( f : D \rightarrow \mathbb{R} \) with \( x_0 \) as an accumulation point of \( D \). Prove that \( f \) has a limit at \( x_0 \) if for each \( \varepsilon > 0 \), there is a neighborhood \( Q \) of \( x_0 \) such that, for any \( x, y \in Q \cap D \), \( x \neq x_0 \), \( y \neq x_0 \), we have \( |f(x) - f(y)| < \varepsilon \).

**2.3 Algebra of Limits**

**16.** Define \( f : (0, 1) \rightarrow \mathbb{R} \) by \( f(x) = \frac{x^3 + 6x^2 + x}{x^2 - 6x} \). Prove that \( f \) has a limit at 0 and find that limit.

**17.** Define \( f : \mathbb{R} \rightarrow \mathbb{R} \) as follows:

\[ 
f(x) = x - [x] \text{ if } [x] \text{ is even.}
\]
\[ 
f(x) = x - [x + 1] \text{ if } [x] \text{ is odd.}
\]

Determine those points where \( f \) has a limit, and justify your conclusions.

**18.** Define \( g : (0, 1) \rightarrow \mathbb{R} \) by \( g(x) = \frac{\sqrt{1 + x} - 1}{x} \). Prove that \( g \) has a limit at 0 and find it.
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Transcribed Image Text:**Chapter 2: Limits of Functions** **15.** Let \( f : D \rightarrow \mathbb{R} \) with \( x_0 \) as an accumulation point of \( D \). Prove that \( f \) has a limit at \( x_0 \) if for each \( \varepsilon > 0 \), there is a neighborhood \( Q \) of \( x_0 \) such that, for any \( x, y \in Q \cap D \), \( x \neq x_0 \), \( y \neq x_0 \), we have \( |f(x) - f(y)| < \varepsilon \). **2.3 Algebra of Limits** **16.** Define \( f : (0, 1) \rightarrow \mathbb{R} \) by \( f(x) = \frac{x^3 + 6x^2 + x}{x^2 - 6x} \). Prove that \( f \) has a limit at 0 and find that limit. **17.** Define \( f : \mathbb{R} \rightarrow \mathbb{R} \) as follows: \[ f(x) = x - [x] \text{ if } [x] \text{ is even.} \] \[ f(x) = x - [x + 1] \text{ if } [x] \text{ is odd.} \] Determine those points where \( f \) has a limit, and justify your conclusions. **18.** Define \( g : (0, 1) \rightarrow \mathbb{R} \) by \( g(x) = \frac{\sqrt{1 + x} - 1}{x} \). Prove that \( g \) has a limit at 0 and find it.
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