Chapter 2: Problem 5
State the precise definition of \(\lim _{x \rightarrow a} f(x)=L\).
Short Answer
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chapter 2: Problem 5
State the precise definition of \(\lim _{x \rightarrow a} f(x)=L\).
These are the key concepts you need to understand to accurately answer the question.
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Get started for freeThe limit at infinity \(\lim _{x \rightarrow \infty} f(x)=L\) means that for any \(\varepsilon>0\) there exists \(N>0\) such that $$|f(x)-L|<\varepsilon \quad \text { whenever } \quad x>N.$$ Use this definition to prove the following statements. $$\lim _{x \rightarrow \infty} \frac{2 x+1}{x}=2$$
Calculate the following limits using the factorization formula \(x^{n}-a^{n}=(x-a)\left(x^{n-1}+x^{n-2} a+x^{n-3} a^{2}+\cdots+x a^{n-2}+a^{n-1}\right)\) where \(n\) is a positive integer and a is a real number. $$\lim _{x \rightarrow 1} \frac{\sqrt[3]{x}-1}{x-1}\left(\text { Hint: } x-1=(\sqrt[3]{x})^{3}-(1)^{3}\text { ). }\right.$$
It can be shown that \(1-\frac{x^{2}}{6} \leq \frac{\sin x}{x} \leq 1,\) for \(x\) near 0 a. Illustrate these inequalities with a graph. b. Use these inequalities to evaluate \(\lim _{x \rightarrow 0} \frac{\sin x}{x}\).
Sketching graphs of functions Sketch the graph of a function with the given properties. You do not need to find a formula for the function. $$\begin{array}{l} g(1)=0, g(2)=1, g(3)=-2, \lim _{x \rightarrow 2} g(x)=0 \\ \lim _{x \rightarrow 3^{-}} g(x)=-1, \lim _{x \rightarrow 3^{+}} g(x)=-2 \end{array}$$
Use the following definitions. Assume fexists for all \(x\) near a with \(x>\) a. We say the limit of \(f(x)\) as \(x\) approaches a from the right of a is \(L\) and write \(\lim _{x \rightarrow a^{+}} f(x)=L,\) if for any \(\varepsilon>0\) there exists \(\delta>0\) such that $$|f(x)-L|<\varepsilon \quad \text { whenever } \quad 0< x-a<\delta$$ Assume fexists for all \(x\) near a with \(x < \) a. We say the limit of \(f(x)\) as \(x\) approaches a from the left of a is \(L\) and write \(\lim _{x \rightarrow a^{-}} f(x)=L,\) if for any \(\varepsilon > 0 \) there exists \(\delta > 0\) such that $$|f(x)-L| < \varepsilon \quad \text { whenever } \quad 0< a-x <\delta$$ Prove that \(\lim _{x \rightarrow 0^{+}} \sqrt{x}=0\).
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