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Find the limit using properties of Iimits. x3 + 9x2 Iim X--9 x2 _ 81[-/1 Points]DETAILSSULLIVANCALC2 1.4.009.Find the limit_ Iim (6x3 + 4 sin x) 1 _[-/1 Points]DETA...

Question

Find the limit using properties of Iimits. x3 + 9x2 Iim X--9 x2 _ 81[-/1 Points]DETAILSSULLIVANCALC2 1.4.009.Find the limit_ Iim (6x3 + 4 sin x) 1 _[-/1 Points]DETAILSSULLIVANCALC2 1.4.011,Find the limit_lim 7J/3cos x + 9 sin x)

Find the limit using properties of Iimits. x3 + 9x2 Iim X--9 x2 _ 81 [-/1 Points] DETAILS SULLIVANCALC2 1.4.009. Find the limit_ Iim (6x3 + 4 sin x) 1 _ [-/1 Points] DETAILS SULLIVANCALC2 1.4.011, Find the limit_ lim 7J/3 cos x + 9 sin x)



Answers

Evaluate the indicated limit, if it exists. Assume that $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$ $$\lim _{x \rightarrow 0} \frac{2 x}{3-\sqrt{x+9}}$$

Okay. So we can start by using our quotient rule to write the falling as the limits of our numerator over denominator. Okay, so in our numerator, we can use our some role, and in our denominator, we can use our constant multiple little. So we commemorate that as limits as execute zero of one plus limits as X approaches zero of X and then plus the limits as X approaches zero of sign of X all over three limits as X approaches zero of co sign of X. So we get one plus zero plus zero over three times co sign of zero, which is one sabbatical to 1/3.

For this problem, we want to find the value of the given limit. We know that the limit in this case is going to be as X approaches three, we're gonna have X squared minus nine Divided by Sine of X -3. Yeah, so in this case, what we have is that the women As X approaches three is going to be based on the fact that because we if we define this as our function F of X F of three is going to be undefined. However, if we Look at this function right after three and right before three, we end up seeing that we'll get a lot closer To our value. So if we have 3.01, we get very close to the value of three. We end up getting that six is our limit. And then if we look at a 2.9 Again six is very close to the limit. So we see that six is going to be the value of our limits.

Using direct substitution immediately. We could replace each of the exes with zero, so this would get US one plus zero plus sign of zero, which is also zero all of that divided by three Times Co sign of zero. However, three times co signing zero would actually give us one. And so, in the end, our final solution here would be 1/3 or 1/3.

Okay. Here have I wanted his ex of purchase Surprise zero The X minus X co sign eggs of here sine squared three x. Okay. I mean, I guess we should notice first that we try to plug in zero. We're going to get zero over zero. No. So let's see what we can do here. Well, the first thing, maybe to try is too down is probably too. Break it up. Okay. What? I mean by that I mean, uh, write this as the limit. It's expert zero uh, x over sine squared three x. Okay. And then subtract the limit is experts zero of eggs cause I necks over sine squared. Re x. Okay, So what can we do here? Well, now this is signs squared. OK? So if I want to be able to I kind of do my usual trick and divine this by X. I actually need to Oh, divide by X squared, actually. Divide by three expert or nine x squared. So this is the limit. It's expertise. Zero x over some of the multiply by nine x squared over nine x squared. Okay. And then minus the limit. The legs coast hero of X because the next time and did the same thing over here with nine x squared sine squared three X over nine x squared and let me make a note here said nine. Ah, I just thought, Why am I doing this? It's a sign squared three X over nine X squared is the same thing as sign three X over three X squared, but this is going to want so therefore, this is going to want is one's greatest one. So what we have Okay, so we have when? Over here. See, this is the limit. So what's what's left after I do this? So X over nine x squared and then minus pat. Okay, coz I nexus going to one That's not a problem. And when x over nine x squared And now Okay, so I'm re combining the limits because actually, the linens excellent. Excellent x squared is going to infinity and then the same thing over here. So I have another indeterminate form, so I need to put him back together, and then these air actually going to zero right before I take the limit to the turtle in it is just zero


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