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Waah. I flunk it so badly, the Asianness in me won't forgive myself lol. The limits I didn't too badly minus the fact that I frigging forgot tanx/x is 1 and limg x-> 0 cosxtanx/x is 1 instead of 0 which the latter was what I put. At least now I can learn from my mistakes, x->0 sinx/x is also 1, right?
Problem one, when you multiply sqrt(x+5) and sqrt(x+5) together, you get x+5, I know that much and -3 * 3 is -9... but what about 3 times sqrt(x+5)?
Problem two, using the derivative rule, I know how to do the basics and such but the fraction bothers me, so it splits? Can someone explain it to me better?
Problem three, please please PLEASE explain to me how;
lim x-> -1^+ g(x) equals 1. Just by looking at the graph, I don't know how to look at this correctly. Someone help with this. I had this on the quiz and I basically do not know what to put as an answer. g(1) I had the vaguest idea that it's answer is -1 because there's a hole there, otherwise... I don't know. Problem three is my utmost important because I don't know how to look at the graph. Thanks in advance.
Edit, problem three has three questions.
 
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for problem 1, (a-b)(a+b)=(a^2-b^2), so it would be x+5-9
for problem 2, i don't know what you're asking, but it's all over h
for problem 3, just follow the curve with your finger. -1+ means -1 from the right hand side, so go from to the left towards -1 and you see it goes to 1
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Sorry, I guess it's a bit sloppy. For problem two, look at the top of the problem two. The bottom one I solved wasn't an issue because it's pretty straight forward, but the top one which is:
lim h->0
f(x+h) - f(x) / h which is the rule.
since it's 1/(2x+3).... it would have to be 1/(2(x+h)+3) - 1/2x+3 OVER h? Okay. I guess I got the rule confused a bit because of the fraction.
for problem 3, just follow the curve with your finger. -1+ means -1 from the right hand side, so go from to the left towards -1 and you see it goes to 1
Okay, so I'm following it now: lim x-> -1^+ g(x) which is -1 on the right hand side and going left, I still don't get it, do you start from the -1? Can you(or someone) draw it via paint? I don't know where to start with it.
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Well you can start anywhere on the graph, as long as it's to the right of -1 and you are going to the left towards -1. For example, start at 0,0 and go left towards -1 and you see that your finger approaches 1.
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Okay, I get it better. for the lim x->1 g(x) though, it comes from both sides, incoming from the left there's a break, continue?
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If it's lim x->1, then it means from both sides, so you go on both sides of the point and move along the graph towards the point and if your fingers meet, that's the limit. your left finger would go towards -2, since you are following the left side of graph, and your right finger would go towards -2, since you are following the right side of the graph, so the limit would be -2. even though the function isn't defined there.
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EPT![[image loading]](http://i29.tinypic.com/672r9y.jpg)
