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Problem 20 Easy Difficulty

Prove the statement using the $ \varepsilon $, $ \delta $ definition of a limit.

$ \displaystyle \lim_{x \to 10}(3 - \frac{4}{5}x) = -5 $


$\delta=\frac{5}{4} \epsilon$


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Video Transcript

This is problem number twenty of the Stuart Calculus eighth edition, Section two point four Prove the statement using the absolute out the definition of a limit. No limit as X approaches ten of the function three minutes for fifty X is equal to negative five and the Epsilon Delta definition of a limit Is that reverie? Absolute Greatest zero. We must find a delta that is created in the room. Ah, such that the following two inequalities are true. One dependent on the other. If the outside value of the difference between X and A is listing delta, then the value of the difference between the function on the limit is less than epsilon. So we are going to use the second inequality start with here and use our function that was given three minus forfeit. Tex, subject our limit. L is equal to taking a five western Absalom and we simplify here combined the like terms three minus thing. If I was eight. Thank you, Mrs Negative for Fitz X plus eight Weston Salaam. One step that we can take is to change this eight into a fraction with denominator of five before five x plus. Already over five that's equal to eight are less than absolute. If we combine, the numerator is here and they get for expose forty. No, that's the numerator, the denominators five. And at this point, we can factor out a negative for fifths. So we're gonna have is the absolute value. Negative forfeits factored out and it leaves us with X plus ten. Sorry, X minus ten. They also have that. Listen, Absalon, this absolutely cancels the negative s. So it's just forfeits. And then we want other five fourths to both sides and that gets us find Absalon or four. So we've gone down to that point. We recall this. First Inequality X minus A is ten in our case in this limit ten less than dull term. And we see the similarities between these two final inequalities and we see that we can find a delta For every Absalon provided that the maximum value that doctor is five absolute or for you can choose anybody. Listen, that. But this is sufficient enough to prove our statement

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