Let f(x) = 4x - 1. To prove that limx → 9 f(x) = 35, we...

Let $f(x) = 4x - 1$. To prove that $\lim_{x \to 9} f(x) = 35$, we proceed as follows. Given any $\epsilon > 0$, we need to find a number $\delta > 0$ such that if $0 < |x - 9| < \delta$, then $|(4x - 1) - 35| < \epsilon$. What is the (largest) choice of $\delta$ that is certain to work? (Your answer will involve $\epsilon$. When entering your answer, type e in place of $\epsilon$.) $\delta = $

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00:01 Okay, so in this question we're asked to use the squeeze theorem to calculate a limit.

00:05 So i'm just going to write everything as it is in the question.

00:09 So we want to use the squeeze theorem.

00:12 And you know what this is? essentially, if a function is between two other functions, and these two other functions converses the same thing as x goes to infinity, for example.

00:23 The one in the middle must also go to the same thing.

00:25 So we want to use the squeeze theorem to show that the limit is x goes to plus infinity.

00:30 Of 1 over x plus e to the minus x is equal to 0.

00:35 And i'm going to put boxes around the things that you should write down.

00:39 So the first thing that we are told is that for x greater equal to 0, because we know that the exponential is always between 0 and 1, we get something for x plus e to the minus x.

00:55 So on the one hand, we know e to the minus x is smaller than 1.

01:00 So this must be smaller than x plus 1.

01:04 So this is what must go on the right.

01:07 And at the same time, e to the minus x is greater than 0, so this must be greater than x plus 0, which is simply x.

01:14 So this is what goes on the left.

01:18 And now we say, therefore, we want something not for x plus e to the minus x, but for its inverse.

01:25 We get something for 1 over x plus e to the minus x...

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