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Numerade Educator



Problem 86 Hard Difficulty

Use Definition 2 directly to prove that $ lim_{n \to \infty} r^n = 0 $ when $ \mid r \mid < 1. $


when $r=0\left|0^{n}\right|<\epsilon$ will always be true


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Video Transcript

let's use definition to to prove that this limit zero Assuming the absolute value. Ours lesson one. So first case one are equal. Zero. There's really nothing to show here. The limit of our end is just a limit of zero the and equal zero. If we really wanted to use the definition for two for case one, it's we can still use it. Let UPS lobby bigger than zero, then just let capital and be one if little and is bigger than capital. And then r n minus zero is less than Epsilon. This is always true because since are equal zero. So our and minus zero is just zero and zero is lost in any positive number. So that's the whole crew for Case one. Now let's look at the more interesting case. So we're still assuming that absolute value. Ours less than one. So let up Salon be bigger than zero. Now let's work backwards. We want little into satisfy this, which is equal to let's say absolute value are the and we want this to be less than Epsilon. So let's just solve this for end. Let's take that log natural log on both sides and then software in noticed the inequality will flip and I'll explain why in one second this is due to the fact that absolute value of ours, less than one so natural log is negative and the absolute value ours non zero. Because foreign case to this is why we have two cases. So now just let capital and the natural lot of e over natural log absolute value are now I'LL go on to the next page. Then if we take a little ends, be bigger than begin it's little and is bigger than begin if then by construction. R n minus zero is Lesson Absalon and by definition, too, this proves the limit of our the end equal zero. If absolute value are is less than one, that's our final answer.