Furry teaches you the Fundamental Limit Laws - Calculus 1

[00:00:00]Hello everyone, MathCat here.

[00:00:03]Let's take a look at some of the fundamental limit laws.

[00:00:06]If I gave you this limit, would you be able to solve it?

[00:00:10]Well, I see a trigonometric function up here, x down here. If we directly plug in zero, that will result in an indeterminate form.

[00:00:21]So, let's try and find a way to simplify this down first before we solve.

[00:00:26]So, we'll need the fundamental limit laws under our belt.

[00:00:32]Before we start, I want to make sure you're familiar with some of the indeterminate forms you might see that we covered in our last limits video. If you're not familiar, I suggest going back and taking a look at that video in order to understand why indeterminate forms do not work for evaluating limits this way.

[00:00:52]Let's say we have the limit as x approaches some constant c. So, it could be any number.

[00:01:02]of f x plus g x.

[00:01:09]This will simplify down to we can kind of distribute this limit notation kind of like a variable, but just so you know, it's not a variable.

[00:01:21]Limit as x approaches c of f of x plus the limit as x approaches c of g of x.

[00:01:38]And let's call this Let's call this l.

[00:01:43]This limit evaluated plus Let's call this m.

[00:01:51]So, you can see here that this limit simply evaluates to calculating the individual limit of each function.

[00:02:02]Let's take a look at some other rules and I'll keep this up here so you remember.

[00:02:09]Um the limit as X approaches some constant C of Let's try the subtraction or the difference rule.

[00:02:26]Can you predict what this will be?

[00:02:29]Well, you're again kind of distributing this limit notation down.

[00:02:34]So, it would be the limit as X approaches C, which is L minus um See how that's kind of uh predictable?

[00:02:48]Now, let's take a look at another form.

[00:02:54]How about the limit as X approaches some constant C of some constant K times a function.

[00:03:12]Here, this would simplify down to calculating the limit of the function and then multiplying it by C.

[00:03:21]So, in this case, it would be K times L.

[00:03:31]Let's take a look at another.

[00:03:35]So, uh let's take a look at the quotient rule.

[00:03:40]The limit as X approaches some constant C again of f of x over g of x.

[00:03:54]This would simplify down to distributing the limit to the top and to the bottom.

[00:04:04]So, it'd be L over M with one constraint.

[00:04:12]M does not equal zero.

[00:04:18]And do you see why?

[00:04:20]Because M equal zero, that would result in a division by zero error, which is undefined.

[00:04:31]Now, let's take a look at maybe some powers.

[00:04:37]The limit as x approaches c, some constant c, of f of x to the power of n.

[00:04:53]Actually, let's use a different color.

[00:04:56]Let's use yellow.

[00:04:58]To the power of n.

[00:05:00]This would evaluate to calculating the limit as x approaches c of this inside function, which would be L.

[00:05:14]And this would be to the power of n.

[00:05:20]So, you evaluate this limit inside first, and this n out here, this is some power um that is unaffected.

[00:05:31]Now, let's take a look at the square root version or the root version.

[00:05:38]x approaches c for the nth root of f of x.

[00:05:52]This would um this would be a special case. You would still get the square root or sorry, the nth root of the function evaluated at the limit. But this comes with a caveat.

[00:06:09]Let's say c was um or sorry, let's say n is three and l evaluated to -8.

[00:06:27]This would simply evaluate to -2 * -2 * -2, right?

[00:06:36]Well, what if I had an even power of say square root of -4.

[00:06:47]Do you see the problem now?

[00:06:50]We have the square root or an even root of a negative number.

[00:06:55]This will result in a imaginary solution or an imaginary number value.

[00:07:03]So, in order for this notation to work, for this rule to work, if n is even l l must be greater or equal than equal to zero.

[00:07:27]If n is odd anything works.

[00:07:38]l can be any value.

[00:07:40]Any real number.

[00:07:44]So, if the limit you see L is um less than or an imaginary number, then you need to be careful. Or sorry, if L is negative, then you need to be careful.

[00:08:00]Because um if n is even, then you can't evaluate that limit that way. The answer doesn't exist.

[00:08:09]D N E Okay.

[00:08:14]The limit as x approaches c Oh, actually uh zero of tangent of x over x The limit rule we'll use for this is is the limit as x approaches some constant c of f of x times g of x.

[00:08:43]This is equal to evaluating the individual limits and multiplying it together. So, we will get L times M Oh, was the L blue? Yeah, the L was blue.

[00:09:00]Um yeah.

[00:09:04]So, knowing that information, would you like to give this limit a shot?

[00:09:10]Hopefully, you tried that. Um let's first try and simplify this down.

[00:09:15]So, we have tangent and you must be familiar with this trigonometric identity.

[00:09:22]sin of x cosine of x So, therefore, we have the limit as x approaches zero of sin x over x times or sorry.

[00:09:44]x times cosine of x and we can kind of separate that cosine away and you'll see why in a little bit.

[00:09:55]So, we have the limit as x approaches zero of sin of x over x times one over cosine of x and do you see here?

[00:10:16]Does this limit look familiar?

[00:10:18]This is the famous squeeze theorem limit.

[00:10:22]This will evaluate to um one and as x approaches zero, cosine equals one or cosine approaches one as x approaches zero.

[00:10:39]So, we have one times one this is due to if you take a look at the unit circle, you'll see.

[00:10:48]Um and we're allowed to do this evaluate the limits of both because of this um fundamental [clears throat] limit rule over here.

[00:10:59]Um we evaluate evaluated each of the limits.

[00:11:04]And therefore, the limit of as x approaches zero of tangent of x over x is equal to one.

[00:11:15]Now, let's take a look at a more conceptual problem.

[00:11:20]All right, I've written out a question for you. If you'd like to try it, actually please do try it, but if not, let's get started. Suppose f of x and g of x are functions defined for all of x near two, and suppose that a and b are some constants.

[00:11:40]All right, so there looks like a lot of writing on the screen right now, but let's break it down and let's start with question one.

[00:11:49]Find the limit as x approaches two of f of x.

[00:11:54]So, um let's pick one of these two to find f of x because remember we're able to distribute this limit notation kind of like a variable, but again, it's not a variable.

[00:12:10]We can distribute that. Um I'll choose this equation.

[00:12:15]That equation simplifies down. Let me use screen.

[00:12:19]The limit as x approaches two of f of x times the limit as x approaches two of g of x.

[00:12:32]This um because we know the limit as x approaches two um for g of x, we can simply substitute two into there.

[00:12:43]And this means this all equals 12. So, um this number must be six.

[00:12:53]So, we know now that the limit as x approaches two of f of x, this this is equal to six.

[00:13:05]Now, we need to determine if this whole thing exists. If so, find the value.

[00:13:13]So, um let's start.

[00:13:18]We can simplify this down using the nth root rule.

[00:13:25]So, we have the square root of the limit as x approaches 2 of f of x times the limit. Let's distribute this limit to the top and to the bottom.

[00:13:44]The limit as x approaches 2 of 1 over the limit as x approaches 2 of g of x.

[00:13:57]And notice here, there's no x in this top variable or top limit, so this just evaluates to 1.

[00:14:07]And now we have the components to solve.

[00:14:11]Because we know g of x is equal to 2 or the limit as x approaches 2 for g of x is equal to 2.

[00:14:20]And this limit evaluates to 6. Just simply plug it in.

[00:14:24]>> [clears throat] >> The square root of 6 * 1/2.

[00:14:30]This is equal to the square root of 6 over 2.

[00:14:33]And this be your answer. I'm sure you can simplify this down.

[00:14:38]Actually, no. This would be your unsimpli- This would be your most simplified answer for number two.

[00:14:47]And it exists because it doesn't result in any division by zero errors or indeterminate forms.

[00:14:55]So, now find a if b is equal to 1.

[00:15:00]We'll need to utilize this top equation up here.

[00:15:05]Um So, yeah. Let's start. Let's start by distributing this limit notation using our constant rule.

[00:15:17]If you remember let's distribute that.

[00:15:20]So, we'll have the limit as X approaches 2 of a * f of x plus the limit as X approaches 2 times b of g of x and that is equal to, if I remember correctly, 10.

[00:15:45]So, we know b is 1. So, that constant doesn't matter. Now, let's just find a.

[00:15:54]Well, we know this evaluates the two in our given equations.

[00:16:00]So, this results in two and then we have we can simply take this constant here and do you remember the product rule where we can distribute the limit to these two terms? Well, if you we distribute the limit to this term, this just evaluates the a.

[00:16:23]And we have the limit as X approaches 2 of f of x.

[00:16:30]And what does that evaluate to?

[00:16:33]Well, we know the limit as X approaches 2 for f of x is equal to 6, right?

[00:16:40]So, this would be a * 6 + 2 is equal to 10.

[00:16:50]Now, um we can just do some simple algebra and we get 6 a is equal to sorry, 8.

[00:17:00]And a is therefore equal to 8/6, which is equal to 4/3.

[00:17:09]And this would be your final answer.

[00:17:12]4/3.

[00:17:16]And that's a wrap for today's lesson.

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[00:18:17]And some updates in regarding future upload schedules.

[00:18:22]Um I'm a student heading to university in a little bit. So, up to uploads may slow down for a little bit, but I'll try and complete this calculus series um over the summer.

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