anonymous
  • anonymous
I'm doing limits and I have a quiz and I don't think I fully get everything. I am struggling to get through the practice problems. lim h->0 [(sqroot 2+h) -2]/h
Calculus1
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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sirm3d
  • sirm3d
like this? \[\Large \lim_{h\rightarrow 0} \frac{\sqrt{2+h}-\sqrt{2}}{h}\]
anonymous
  • anonymous
just rationalize the numerator
anonymous
  • anonymous
and then take the limits after cancelling out the common factor

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anonymous
  • anonymous
@sirm3d the -2 is not \[\sqrt{2}\]. so the numerator is \[\sqrt{2+h}-2\]
anonymous
  • anonymous
@cali2 check ur question if it is as sirm3d has written
anonymous
  • anonymous
and if u r correct then the limit is undefined
sirm3d
  • sirm3d
oh. then we're looking at an infinite limit, or that the limit does not exist.
anonymous
  • anonymous
@sirm3d or @matricked do u mind explaining how you arrived at that?
calculusfunctions
  • calculusfunctions
Since substitution yields 0/0 (an indeterminate form), we rationalize the numerator by multiplying both the numerator and denominator by the conjugate of the numerator, because there is a square root function involved. Do you know how to proceed from here?
anonymous
  • anonymous
@calculusfunctions I got to that part and further. But I can't get further than here, when i did i got \[1/\sqrt{2}\] so I am not sure where I did something wrong. \[\lim_{h \rightarrow 0} 2+h -4/h \sqrt{2+h}+2\]
calculusfunctions
  • calculusfunctions
OK fine, I'll walk you through a few more steps only because it seems like you've made the effort, and are not simply looking for someone to just give you the solution.\[\lim_{h \rightarrow 0}\frac{ \sqrt{2+h}-\sqrt{2} }{ h }⋅\frac{ \sqrt{2+h}+\sqrt{2} }{ \sqrt{2+h}+\sqrt{2} }\]\[=\lim_{h \rightarrow 0}\frac{ 2+h -2 }{ h(\sqrt{2+h}+\sqrt{2}) }\]\[=\lim_{h \rightarrow 0}\frac{ h }{ h(\sqrt{2+h}+\sqrt{2}) }\]\[=\lim_{h \rightarrow 0}\frac{ 1 }{ \sqrt{2+h}+\sqrt{2} }\]@cali2 do you see? You wrote the question incorrectly. It should be minus root 2, not minus 2.
anonymous
  • anonymous
The question from the book just has it as minus 2 to begin with. Is there somewhere I need to be changing it? Also, if it were to be minus root 2 and the way I have done minus 2 as well, don't you get a value? I guess that is what I don't understand, I thought after trying all other options apart from substitution and then substitution and you don't get a value is when you can conclude it doesn't exist. Or is it that after working that out I have to work out the one sided limits. Is that wrong?
calculusfunctions
  • calculusfunctions
Well, either your textbook has inadvertently misprinted the question, and it should be minus root 2, or if the textbook question is correct then the answer should be "does not exist", or the question should have root(4 + h) - 2 in the numerator. There are several plausible scenarios but either way the textbook seems to be at fault here, unless you didn't read the question carefully.
anonymous
  • anonymous
The answer in the back of the book says that the limit does not exist. But I do not understand why
calculusfunctions
  • calculusfunctions
That's one of the scenario's I stated above. Then the question is written correctly! By substituting h = 0 into the given function, we see that the numerator is (root 2) -2 and the denominator is 0. Thus, any non-zero value divided by zero is automatically undefined, and therefore the limit does not exist.
anonymous
  • anonymous
Oh okay. So I would just need to stop there. I thought I had to go through the whole conjugation thing. I think I get it. I assume conjugation/factorization only needs to happen if I get 0/0 and I can't factor either
calculusfunctions
  • calculusfunctions
Yes you only proceed with algebraic manipulation if the "limit" is an indeterminate form.
anonymous
  • anonymous
okay, THANK YOU!
calculusfunctions
  • calculusfunctions
Welcome! Glad you understand.
calculusfunctions
  • calculusfunctions
Do you need help with anything else?
calculusfunctions
  • calculusfunctions
I have to go now so good luck!
anonymous
  • anonymous
Actually yes. I do not know how to approach these. I can go further in #2 but I have no idea how to even start #1. Both these limits are \[\lim_{x \rightarrow \Pi/4}\] 1. \[\frac{ \sin x-\cos x }{\tan x-1}\] 2. \[(\frac{ 1 }{ \tan x -1}-\frac{ 2 }{ \tan ^{2}x-1 })\] And for this one I got 4 as the answer but the answer is 2 \[\lim_{x \rightarrow 4}\frac{ x-4 }{ \sqrt{x}-\sqrt{8-x} }\] I did the conjugate here but my answer is 4.
anonymous
  • anonymous
ok, well thanks anyway! Appreciate it
calculusfunctions
  • calculusfunctions
@cali2 sorry I didn't know how much longer you'd be so I had to leave. For #2, Simply add the two fractions and the rest will work itself out. For that last one the answer is 2, so the textbook is correct. Try it again. I'll check in later.
phi
  • phi
For the trig questions, try changing everything into sin or cos for the last problem \[ \lim_{x \rightarrow 4}\frac{ x-4 }{ \sqrt{x}-\sqrt{8-x} } \] multiply by the conjugate \[ \lim_{x \rightarrow 4}\frac{ x-4 }{ \sqrt{x}-\sqrt{8-x} } \cdot \frac { \sqrt{x}+\sqrt{8-x}}{ \sqrt{x}+\sqrt{8-x}} \] be careful with the signs, but the bottom is \[ \lim_{x \rightarrow 4}\frac{( x-4)(\sqrt{x}+\sqrt{8-x}) }{ x-(8-x)} \] or \[ \lim_{x \rightarrow 4}\frac{( x-4)(\sqrt{x}+\sqrt{8-x}) }{ 2(x-4)} \] now cancel the (x-4) and let x -> 4

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