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

- anonymous

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- sirm3d

like this?
\[\Large \lim_{h\rightarrow 0} \frac{\sqrt{2+h}-\sqrt{2}}{h}\]

- anonymous

just rationalize the numerator

- anonymous

and then take the limits after cancelling out the common factor

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## More answers

- anonymous

@sirm3d the -2 is not \[\sqrt{2}\]. so the numerator is \[\sqrt{2+h}-2\]

- anonymous

@cali2
check ur question if it is as sirm3d has written

- anonymous

and if u r correct then the limit is undefined

- sirm3d

oh. then we're looking at an infinite limit, or that the limit does not exist.

- anonymous

@sirm3d or @matricked do u mind explaining how you arrived at that?

- 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

@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

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

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

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

The answer in the back of the book says that the limit does not exist. But I do not understand why

- 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

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

Yes you only proceed with algebraic manipulation if the "limit" is an indeterminate form.

- anonymous

okay, THANK YOU!

- calculusfunctions

Welcome! Glad you understand.

- calculusfunctions

Do you need help with anything else?

- calculusfunctions

I have to go now so good luck!

- 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

ok, well thanks anyway! Appreciate it

- 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

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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