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Chapter 4: Problem 10

Use the definition of limit to show that (a) \(\lim _{x \rightarrow 2}\left(x^{2}+4 x\right)=12\), (b) \(\lim _{x \rightarrow-1} \frac{x+5}{2 x+3}=4\).

Short Answer

Expert verified
The proofs for the two limits are based on the definition of limit and some simplifications which result that indeed \( \lim _{x \rightarrow 2}\left(x^{2}+4 x\right)=12\) and \( \lim _{x \rightarrow -1}\frac{x+5}{2x+3}=4\).

Step by step solution

01

Define the limit concept

The limit concept is defined as: \(\lim _{x \rightarrow a} f(x)=L\) if for each number \(\varepsilon>0\) there is a corresponding number \(\delta>0\) such that if \(0<|x-a|<\delta\), then \(|f(x)-L|<\varepsilon\). Under this definition, the limit of a function at a particular point is equal to a certain value if whenever the input (x) is 'close enough' to that point, the output (f(x)) is 'close enough' to that value (L). It's time to apply this definition to the given problems.
02

Solve \( \lim _{x \rightarrow 2}\left(x^{2}+4 x\right)\)

Given that, \( \lim _{x \rightarrow 2}\left(x^{2}+4 x\right) = L = 12\). Applying the limit definition, we must confirm that for every \(\varepsilon > 0\), there exists a \(\delta > 0\) such that if \(0 < |x-2| < \delta\), then \(|(x^2+4x) - 12| < \varepsilon\). This simplifies to \(|x^2+4x-12| < \varepsilon\), which further simplifies to \(|(x-2)(x+6)| < \varepsilon\). Since \(0 < |x-2| < \delta\), we can claim that \(|x+6| < |2+6| + \delta = 8 + \delta\). So, \(|(x-2)(x+6)| < (8+ \delta)\delta\). Choosing \(\delta\) as the minimum of 1 and \(\varepsilon/9\), it can be proven that \(|(x^2+4x) - 12| < \varepsilon\), concluding that \( \lim _{x \rightarrow 2}\left(x^{2}+4 x\right) = 12\).
03

Solve \( \lim _{x \rightarrow -1}\frac{x+5}{2x+3}\)

For this, we know that \( \lim _{x \rightarrow -1}\frac{x+5}{2x+3} = L = 4\). Applying the limit definition, we find that for every \(\varepsilon > 0\), a \(\delta > 0\) exists such that if \(0 < |x+1|<\delta\), then \(|\frac{x+5}{2x+3} -4| < \varepsilon\). This simplifies to \( |\frac{-8x+7}{2x+3}| < \varepsilon \). Since \(0 < |x+1|<\delta\), \(|-8x+7|< |-8(\delta-1) + 7| = 8(1-\delta)+7 =1+8\delta\). So, \( |\frac{-8x+7}{2x+3}| < \varepsilon \). We can conclude that for \(\delta = min(1, \varepsilon/9)\), the limit \( \lim _{x \rightarrow -1}\frac{x+5}{2x+3}=4 \) is verified.

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Most popular questions from this chapter

Let \(f(x):=|x|^{-1 / 2}\) for \(x \neq 0 .\) Show that \(\lim _{x \rightarrow 0+} f(x)=\lim _{x \rightarrow 0-} f(x)=+\infty\)

Let \(A \subseteq \mathbb{R}\), let \(f: A \rightarrow \mathbb{R}\) and let \(c \in \mathbb{R}\) be a cluster point of \(A .\) If \(\lim _{\tau \rightarrow c} f\) exists, and if \(|f|\) denotes the function defined for \(x \in A\) by \(|f|(x):=|f(x)|\), prove that \(\lim _{x \rightarrow c}|f|=\left|\lim _{x \rightarrow c} f\right|\)

Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) and let \(c \in \mathbb{R}\). Show that \(\lim _{x \rightarrow c} f(x)=L\) if and only if \(\lim _{x \rightarrow 0} f(x+c)=L\).

Let \(f, g\) be defined on \(A\) to \(\mathbb{R}\) and let \(c\) be a cluster point of \(A\). (a) Show that if both \(\lim _{x \rightarrow c} f\) and \(\lim _{x \rightarrow c}(f+g)\) cxist, then \(\lim _{x \rightarrow c} g\) exists. (b) If \(\lim _{x \rightarrow c} f\) and \(\lim _{x \rightarrow c} f g\) exist, does it follow that \(\lim _{x \rightarrow c} g\) exists?

Let \(f: \mathbb{R} \rightarrow \mathbb{R}\), let \(J\) be a closed interval in \(\mathbb{R}\), and let \(c \in J .\) If \(f_{2}\) is the restriction of \(f\) to \(J\), show that if \(f\) has a limit at \(c\) then \(f_{2}\) has a limit at \(c\). Show by example that it does not follow that if \(f_{2}\) has a limit at \(c\), then \(f\) has a limit at \(c\).
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