Section 2.6 Homework (continued) end behavior of the following function: (Do this algebraically Determine theN16x8 12x2 7x2 + 5/x"o +2x _ 8C. Find the limit or...

Use 1'Hopital's Rule to find the limits in Exercises $7-26$.
$$
\lim _{x \rightarrow \pi / 2} \frac{1-\sin x}{1+\cos 2 x}
$$

Just a friendly reminder for low P tells rule we wanna have some sort of indeterminant form of 0/0 or infinity over infinity. Um, I'm pretty sure it's gonna be 0/0 of two differential functions. So a zoo look at sign of X minus co sin of X, that is differential. So we're probably going to apply Low P tells rule Onda the same thing with X minus pi before, uh, and we're doing the limit as X approaches power before. Yeah, so just a friendly reminder is on the unit circle PIRA four is the ordered pair route to over to route to Over to Yeah, So as I look at that, uh, the other thing I should say is you don't want to write equals 0/0 because that is a false statement that the answer is not equal to that. Which is why a lot of math teachers will ask their students, right arrows, um, like, sign of pirate four because they're doing direct substitution route to over two. And then coastlines also routes over two. Because it's the same x y coordinate Forgot to write me. That equals zero. So I'm showing that the numerator is zero. And if I do the denominator direct substitution again gives me zero eso do not write this. Teachers don't want to see that they want to see arrows and then say, Oh, well, we have 0/0 so we can apply loopy tiles rules. So what is Loki tells rule is, that's the drift of of the top switch colors. Uh, we'll just rewrite this when his ex approaches pirate for just do the derivative of the top. That drift of the signs co sign and then the derivative of negative co sign is positive sign. Because the derivative of coastline is negative signs, it's got changed. The sign, and then a drift of the bottom George of of excess one and a derivative of a constant power for is nothing, so there's no need to write that down. So now direct substitution says Okay, well, co signs route to over to sign is also route to over two. We already talked about that, so denominators are the same. One or two plus one or two is to root two, and then you can simplify that to be just route to your correct answer

In Problem 82. We want to evaluate this limit when X approaches zero graphically New American and analytics. Let's start by the graphical method. We will use this most to grab the function f of X and let's zoom in about X equals zero. When we approach the function from the right, we can see that we can get the value off minus 4.25 And when we approach it from left, we can see that we give the value off minus 4.25 minus one quarter. Then, graphically, the value is minus one quarter. It's get it numerically. When we approach X equals zero, we will construct two tables the first table toe approach, the value from the left. Like these points, we can see that when we approach from the left, the value goes toe minus minus 4.25 here and when we approach from the right For the second list, we can see that the venue goes toe minus 4.25 which is this very then, graphically and numerically, the limit equals minus one part. Let's do it. And politically, let's substitute by X equals zero first when we substitute, we get the value off one minus one, divided by zero, which is zero divided by zero. Then we can a ploy number tells Rule, then it equals the limit off the differentiation off denominator, we differentiate in a minute. The definition off sine X is minus sign x. The differentiation of minus one is just here, divided by the differentiation. Off two X squared is four x When x approaches you, let's substitute again. By substitution, we get minus zero divided by zero, which is zero divided by zero. Again, let's apply. The new Beatles rule again equals the limit. Oh, the differentiation off the name NATO We differentiate Sine X, which he gives cosigned X. And then that's differentiate the denominator differentiation off for just four when X approaches zero. Let's substitute for the third time by X equals zero. We have minus one divided by four, which is the same answer we've got from the graphical and numerical methods.

To find the limit of sign of seven X over X. As X approaches zero, we know that the limit of sine of theta over say to as a state approaches zero, this is equal to one. And so to be able to use this formula, we want the angle and the denominator to be the same. So because we have sign of seven X over X. To make the denominator seven, we have to multiply this by 7/7 and so and here we have limit as X approaches zero of seven times sine of seven X over seven X. And this is equal to limit as X approaches zero of seven times limit as X approaches zero of sign of seven X over seven X. No since seven is a constant, then the first limit would be seven. This times the limit of sign of seven X over seven X, which is one. And so we have seven as the value of the limits.

This question as what is the limit of exploits? Sine X divided by X as X approaches zero and to support a graphically at first and then to confirm algebraic lee so graphically when we graviton a graphing calculator we can see the ass X approaches zero from both the left and the right it approaches to therefore graphically we know that the limit is two i'll go break Lee we have to look, we have to consider this at first. So on page sixty it says that, you know, figured to point one as well Is the graph the table for the graph? It says that the limit of sine X divide by ex as except put zero is one will keep this in mind as we saw the algebraic Lee So I'll just break Lee. I have a set up right here just rewriting the problem. We can split this into two different limits using the sum rule. It's the first part that we're going to split it up to is limit of X as experts zero when the ex Div ibex and then the second limit is limit of sine X, divided by X as X approaches zero. Therefore, we can make this a little simpler to solve. What is extra vibe X We'LL just be one So we as the limit of one as acts of push zero plus limit of sine X buybacks as X approaches. Zero. What's the limit? Here is just one, because there's no X for us to plug into so the limit of one as X approaches. Zero always be just one. And we know that from pay sixty with this graph proof right here that the limit of signing survival X as X approaches zero is also one. Therefore, one plus one is two. We can confirm that algebraic lee the limit of X plus i necks divided by X ask x approaches zero is also too.