Suppose lim⁡x→a f(x)=L{\displaystyle\lim_{x\to a}}\text{ }f\left(x\right)=Lx→alim f(x)=L. Prove that lim⁡x→a (c(f(x))=cL{\displaystyle\lim_{x\to a}}\text{ (}c\left(f\left(x\right)\right)=cL, where cc is a constant.

Start by recalling the definition of a limit: $$ \lim_{x \to a} f(x) = L $$ means that for every $$ \epsilon \u003e 0 $$, there exists a $$ \delta \u003e 0 $$ such that if $$ 0 0 $$, there exists a $$ \delta \u003e 0 $$ such that if $$ 0 0 $$, there exists a $$ \delta \u003e 0 $$ such that if $$ 0 0 $$ such that if $$ 0 \u003c |x - a| \u003c \delta $$, then $$ |c \cdot f(x) - cL| = |c| \cdot |f(x) - L| \u003c |c| \cdot \frac{\epsilon}{|c|} = \epsilon . $$This completes the proof that $$ \lim_{x \to a} (c \cdot f(x)) = cL .$$

Ch. 2 - Limits

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 2 - Limits

Problem 2.7.44

Chapter 2, Problem 2.7.44

Suppose ${$\displaystyle\]\lim$_{x$\to$ a}}$\text{ }$f$\left$(x$\right$)=L$ . Prove that $\lim_{x \to a} (c (f (x)) = c L$ , where $c$ is a constant.

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Start by recalling the definition of a limit: $ \lim_{x \to a} f(x) = L $ means that for every $ \epsilon > 0 $, there exists a $ \delta > 0 $ such that if $ 0 < |x - a| < \delta $, then $ |f(x) - L| < \epsilon $.

To prove $ \lim_{x \to a} (c \cdot f(x)) = cL $, we need to show that for every $ \epsilon > 0 $, there exists a $ \delta > 0 $ such that if $ 0 < |x - a| < \delta $, then $ |c \cdot f(x) - cL| < \epsilon $.

Notice that $ |c \cdot f(x) - cL| = |c| \cdot |f(x) - L| $. We can use the property of absolute values that $ |c| \cdot |f(x) - L| < \epsilon $ if $ |f(x) - L| < \frac{\epsilon}{|c|} $.

Since $ \lim_{x \to a} f(x) = L $, for every $ \epsilon' > 0 $, there exists a $ \delta > 0 $ such that if $ 0 < |x - a| < \delta $, then $ |f(x) - L| < \epsilon' $. Choose $ \epsilon' = \frac{\epsilon}{|c|} $.

Thus, for this choice of $ \epsilon' $, there exists a $ \delta > 0 $ such that if $ 0 < |x - a| < \delta $, then $ |c \cdot f(x) - cL| = |c| \cdot |f(x) - L| < |c| \cdot \frac{\epsilon}{|c|} = \epsilon $. This completes the proof that $ \lim_{x \to a} (c \cdot f(x)) = cL $.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Limit of a Function

The limit of a function describes the behavior of the function as the input approaches a certain value. Formally, we say that the limit of f(x) as x approaches a is L if, for every small positive number ε, there exists a corresponding small positive number δ such that whenever 0 < |x - a| < δ, it follows that |f(x) - L| < ε. This concept is fundamental in calculus as it lays the groundwork for continuity and differentiability.

Constant Multiple Rule

The Constant Multiple Rule states that if a function f(x) approaches a limit L as x approaches a, then the function c*f(x) approaches c*L, where c is a constant. This rule is essential for manipulating limits involving constants and is often used in proofs and calculations to simplify expressions while maintaining their limit properties.

Proof Techniques in Calculus

Proof techniques in calculus often involve epsilon-delta arguments, direct substitution, or algebraic manipulation to establish the validity of limit statements. In the context of limits, proving that the limit of a constant multiplied by a function equals the constant multiplied by the limit of the function requires careful application of these techniques to ensure that all conditions of the limit definition are satisfied.

Suppose limx→a f(x)=L{\(\displaystyle\]\lim\)_{x\(\to\) a}}\(\text{ }\)f\(\left\)(x\(\right\))=Lx→alim f(x)=L. Prove that limx→a (c(f(x))=cL{\(\displaystyle\]\lim\)_{x\(\to\) a}}\(\text{ (}\)c\(\left\)(f\(\left\)(x\(\right\))\(\right\))=cL, where cc is a constant.

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Key Concepts

Limit of a Function

Constant Multiple Rule

Proof Techniques in Calculus

Suppose ${\(\displaystyle\]\lim\)_{x\(\to\) a}}\(\text{ }\)f\(\left\)(x\(\right\))=L$ . Prove that $\lim_{x \to a} (c (f (x)) = c L$ , where $c$ is a constant.