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cali2

  • one year ago

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

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  1. sirm3d
    • one year ago
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    like this? \[\Large \lim_{h\rightarrow 0} \frac{\sqrt{2+h}-\sqrt{2}}{h}\]

  2. matricked
    • one year ago
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    just rationalize the numerator

  3. matricked
    • one year ago
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    and then take the limits after cancelling out the common factor

  4. cali2
    • one year ago
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    @sirm3d the -2 is not \[\sqrt{2}\]. so the numerator is \[\sqrt{2+h}-2\]

  5. matricked
    • one year ago
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    @cali2 check ur question if it is as sirm3d has written

  6. matricked
    • one year ago
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    and if u r correct then the limit is undefined

  7. sirm3d
    • one year ago
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    oh. then we're looking at an infinite limit, or that the limit does not exist.

  8. cali2
    • one year ago
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    @sirm3d or @matricked do u mind explaining how you arrived at that?

  9. calculusfunctions
    • one year ago
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    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?

  10. cali2
    • one year ago
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    @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. \[\lim_{h \rightarrow 0} 2+h -4/h \sqrt{2+h}+2\]

  11. calculusfunctions
    • one year ago
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    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.

  12. cali2
    • one year ago
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    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?

  13. calculusfunctions
    • one year ago
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    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.

  14. cali2
    • one year ago
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    The answer in the back of the book says that the limit does not exist. But I do not understand why

  15. calculusfunctions
    • one year ago
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    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.

  16. cali2
    • one year ago
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    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

  17. calculusfunctions
    • one year ago
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    Yes you only proceed with algebraic manipulation if the "limit" is an indeterminate form.

  18. cali2
    • one year ago
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    okay, THANK YOU!

  19. calculusfunctions
    • one year ago
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    Welcome! Glad you understand.

  20. calculusfunctions
    • one year ago
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    Do you need help with anything else?

  21. calculusfunctions
    • one year ago
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    I have to go now so good luck!

  22. cali2
    • one year ago
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    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.

  23. cali2
    • one year ago
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    ok, well thanks anyway! Appreciate it

  24. calculusfunctions
    • one year ago
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    @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.

  25. phi
    • one year ago
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    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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