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

Problema 2.7.44

Capitolo 2, Problema 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.

Guida verificata passo dopo passo

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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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.

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