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# Prove the statement using the $\varepsilon$, $\delta$ definition of a limit and illustrate with a diagram like Figure 9.$\displaystyle \lim_{x \to -3}(1 - 4x) = 13$

## Given $e>0,$ we need $\delta>0$ such that if $0<|x-(-3)|<\delta$, then$\begin{array}{l}|(1-4 x)-13|<\varepsilon . \text { But }|(1-4 x)-13|<\varepsilon \Leftrightarrow \\|-4 x-12|<\epsilon \Leftrightarrow|-4||x+3|<\epsilon \Leftrightarrow|x-(-3)|<\varepsilon / 4 \text { . So if }\end{array}$we choose $\delta=\varepsilon / 4,$ then $0<|x-(-3)|<\delta \Rightarrow|(1-4 x)-13|<\varepsilon$Thus, $\lim _{x \rightarrow-3}(1-4 x)=13$ by the definition of a limit.

Limits

Derivatives

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##### Top Calculus 1 / AB Educators   ##### Samuel H.

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Idaho State University

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### Video Transcript

So on this problem were asked to prove the limit. As X approaches let's make that a kleenex there approaches minus three. Right? Of 1 -4 x equals 13. Using the epsilon delta definition. Mhm Okay. So by the epsilon delta definition, what? This says that for all epsilon greater than zero, there exists a delta. So that when x minus A is less than delta, we have f of x minus L less than epsilon. Right? All right. So let epsilon created in zero be given. Okay then A is minus three. Um F of X is one month for X and L is 13. All right. So then choose delta To be absolute over four. Uh huh. So then X no -A -3. Right, less than delta. Okay, no means survive X plus three is less than delta. Yeah. Okay. Yes. Then we have Absolute value of 1 -4 X. Let's see, I mean let me let me write it this way for us. Perhaps the value of F of x minus L is well f of X is one minus four X. That's one minus four. X -L. is 13. Okay, so that is absolute value of minus 12 minus four X. Which Is absolutely of -4x. Because like a factor I'm -4. Excuse me. Out of that. So I have three plus x. Okay, What's the value of -4? Right, So this is four, Three months three plus X is less than delta. So I need to say that this is now less than for delta. The delta is epsilon over four. so this is four times epsilon over four, so equals epsilon. Thus The limit, as x approaches -3, uh 1 -4 X is 13, rather by the epsilon delta definition.

DM
Oklahoma State University

#### Topics

Limits

Derivatives

##### Top Calculus 1 / AB Educators   ##### Samuel H.

University of Nottingham ##### Michael J.

Idaho State University

Lectures

Join Bootcamp