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Maurinko [17]
3 years ago
13

Please check screenshot

Mathematics
1 answer:
nata0808 [166]3 years ago
6 0

Answer:

im sorry but whats the question exactly i will try to awnser it

Step-by-step explanation

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babunello [35]

Answer:

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hope this helps

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Step-by-step explanation:

5 0
3 years ago
Let f (x) = 3x − 1 and ε &gt; 0. Find a δ &gt; 0 such that 0 &lt; ∣x − 5∣ &lt; δ implies ∣f (x) − 14∣ &lt; ε. (Find the largest
s344n2d4d5 [400]

Answer:

This proves that f is continous at x=5.

Step-by-step explanation:

Taking f(x) = 3x-1 and \varepsilon>0, we want to find a \delta such that |f(x)-14|

At first, we will assume that this delta exists and we will try to figure out its value.

Suppose that |x-5|. Then

|f(x)-14| = |3x-1-14| = |3x-15|=|3(x-5)| = 3|x-5|< 3\delta.

Then, if |x-5|, then |f(x)-14|. So, in this case, if 3\delta \leq \varepsilon we get that |f(x)-14|. The maximum value of delta is \frac{\varepsilon}{3}.

By definition, this procedure proves that \lim_{x\to 5}f(x) = 14. Note that f(5)=14, so this proves that f is continous at x=5.

3 0
3 years ago
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