Given limx to infty ln15 13 x use math1044 convention for...

Given $\lim_{x \to \infty} \ln(15 + 1.3^{-x})$. Use MATH1044 Convention for Numerical Estimation to numerically estimate the limits to three decimal places. Put NA on all rows that are not needed and/or don't have a value according the the convention. $x \to \infty$ $\ln(15 + 1.3^{-x})$ 10 2.713 100 2.71 1000 2.71 10000 2.71 100000 2.71 Hence $\lim_{x \to \infty} \ln(15 + 1.3^{-x}) = 2.71$

Transcript

00:01 All right, so we're given the limit.

00:03 Limit is x becomes infinite of the natural log of 15 plus 1 .3 to the negative x, which we could write this 1 .3 to the negative x as 1 over 1 .3 to the x.

00:25 Either way, the same thing.

00:26 So what we're doing is we're going to use, we're told you use a specific convention for numerical estimation to estimate this limit.

00:37 But the prompt doesn't say what that is.

00:40 So we're going to show you what the values would be, regardless of whatever estimation method you use.

00:46 So essentially what you're doing is we want to figure out what value does this function approach as x becomes massive.

00:56 Huge.

00:58 Huge, huge.

01:00 So if we think, about this, if we just let x be, just let x go get huge.

01:08 So that means you have natural log of 15 plus 1 over 1 .3 to some huge number.

01:19 We'll say, i mean, you wouldn't write this, but well, let's write it out.

01:23 I was going to write infinity like this, but some mathematicians would call that abusive notation.

01:30 So let's just say it's a big number, 1 .00, or 1.

01:34 Not one point, but one with a bunch of zeros, like lots of zeros.

01:38 So what's going to happen is this term here, this is going to approach zero as this x value becomes infinite.

01:46 And so what you're left with is the natural log of 15.

01:49 So as x becomes massive, this function will approach natural log of 15.

01:55 You could also take this function, if we call this, let's call this natural log of 15 plus 1 .3 to negative.

02:04 Let's just call this f of x.

02:06 What we could do is we could evaluate f of f at different values...

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