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Sin(5h) Evaluate: lim h+0 5h...

Question

Sin(5h) Evaluate: lim h+0 5h

sin(5h) Evaluate: lim h+0 5h



Answers

Evaluate the limit, if it exists.

$ \displaystyle \lim_{h \to 0}\frac{(-5 + h)^2 - 25}{h} $

Limits. H approaches zero of negative five plus H squared rate a minus 25 over H. There are many ways you can do this one, right? You can use C C A swear minus b squared. This. You can consider this when you're a you're just one is gonna be be square, which is five squared. So these five and then you can do a lot of things. There are many things you can do, right? But I'm gonna choose to expand those parent disease so expanding it is gonna be 25 minus, twice times, five times. So 10 h and then close H squared, then minus 25 again over age. So there's 25 minutes 25 goes away, right? So what am I gonna get, Greg? Its limits. It tends to zero negative. 10 h plus h squared. Right. So once I have that, then what I'm gonna do is over h to navigate this one, and then I'm gonna factor that I'm gonna do factor. So this is a have a negative 10 plus h and then over H. And this h slash is this age, so I just have a limit. Each approaches zero off. Negative 10 plus h. So if h goes to zero, then it probably getting that you're gonna have is negative 10.

This limit that was given in this section to tryto work through this result, um, as you can probably imagine. But it's tempting to look at this and try to think, Well, how come you get a five in here along with the agents of the sine function to make this work out? Well, it's actually not really meant for that. In fact, you can do it in a different way entirely. That doesn't involve you trying to get him something inside this, in fact, that really won't work at all. Um, in fact, it's everything that you really need is right in front of you. Um think of this fraction here as instead, um, sign of H over H and it's being multiplied by a 1/5 out front. I think of it that way. Well, by using the limit laws that we learned earlier in the Champion section three, uh, this can simply were rewritten as this 1/5 is the coefficient out front and pulled it all way out front. And then you got the limit of the remaining portion. His age approaches zero of and you've got similar expression, which you have over here in the corner going on now we have sign of H over H and this just like the corner purchase one. So overall we have 1/5 outfront multiplied by them. One received from this limit Give us an overall result of 1/5 for this problem.

So here we have a specific limit. The limit as X approaches five of x squared minus six. X plus five Divided by X -5. So the first thing we can check if this is an indeterminate form and we can evaluate the limit. So 5 -5 is zero. And this we can also find is also equivalent. 20. So as a result we have an indeterminate form and we can value to limit. So one thing we can do is first to factor the numerator. So this would be X -5 Times X -1 Over X -5. So we can cross out the X -5 to simplify down our expression and we have this remaining. So if we can use direct substitution here, since we don't have an indeterminate form And we find that this is equivalent to four. An alternative way of doing this would be applying logical rule. So applying law in the world would also give us uh the same answer. However, a lot the world would require a bit more work as we have to differentiate the numerator and denominator. And but this gives our final answer and the limits equivalent to four

So in this problem were asked to evaluate the limit As X approaches five of this function X squared minus five. X plus six. All over x minus five. Well, we noticed first of all that in the denominator, If we evaluate X -5 at five we get zero and you can't divide I zero. So x equals five is a vertical. Ask them to. Therefore this limit does not exist and we can see this if we graph it for a minute, if we go to a graphing calculator and enter the formula in and graph it, you see this right And x equals five. You notice that it's a vertical assam vote? There is no value of this function on on this equation. X equals five. Right? Excellent. five. There is no value of the function of the red function on that vertical aspect to it right there. So it was. So the limit, you notice that limit from the left goes to negative infinity and live from the right goes to positive infinity. As we approach from the left or the right, depending on which one you're looking at and those are not the same number. And so therefore this limit does not exist


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