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Calculus and L'Hospital's Rule (help me)
ceres
When the last moon is cast over the last star of morningAnd the future has past without even a last desperate warningRegistered User, Moderator Mod Emeritus
When the last moon is cast over the last star of morningAnd the future has past without even a last desperate warningRegistered User, Moderator Mod Emeritus
So I have these math problems, and they aren't going well. The answers are given in the back of the book.
The first is the limit as x approaches infinity of xsin(pi/x).
For this one, I tried making it
x/(1/sin(pi/x))
for infinity/infinity, and then got stuck on differentiating. I came up with
1/(-csc(pi/x) cot(pi/x))
and now I'm just... stuck. The answer in the back of the book is "pi". I just have no freaking clue how to get there.
The next L'Hospital's Rule problem is the limit as x approaches zero of (1-2x)^(1/x).
The first thing I tried was
(1/x)ln(1-2x)
for
ln(1-2x)/x
but as far as I can tell this gives you zero/infinity, and I'm not sure you can even use L'Hospital's Rule for that. Can I differentiate this using the rule as it is?
Help?
The first is the limit as x approaches infinity of xsin(pi/x).
For this one, I tried making it
x/(1/sin(pi/x))
for infinity/infinity, and then got stuck on differentiating. I came up with
1/(-csc(pi/x) cot(pi/x))
and now I'm just... stuck. The answer in the back of the book is "pi". I just have no freaking clue how to get there.
The next L'Hospital's Rule problem is the limit as x approaches zero of (1-2x)^(1/x).
The first thing I tried was
(1/x)ln(1-2x)
for
ln(1-2x)/x
but as far as I can tell this gives you zero/infinity, and I'm not sure you can even use L'Hospital's Rule for that. Can I differentiate this using the rule as it is?
Help?
And it seems like all is dying, and would leave the world to mourn
ceres on
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Posts
You differentiated the denominator incorrectly, but I wouldn't do that anyway. I'd write it as sin(pi/x) / (1/x). Applying L'Hospital's rule, you take the limit of pi*cos(pi/x)*(-x^-2) / (-x^-2) = pi*cos(pi/x) -> pi.
No, it gives you -Infinity/Infinity. Apply L'Hospital's Rule for (1/1-2x)*-2 / 1 -> 0
Well, the typical method to do limits of the form 0^0 is to exponentiate the logarithm so lim(x->infinity) (1-2x)^(1/x) would turn into lim(x->infinity) e^((1/x) ln (1-2x)).
However, ln(1 - 2x) isn't defined for x >= .5, so you may want to make sure that you wrote the problem down correctly. Throwing in roots of negative numbers, which is what will happen with the problem as stated, really complicates things.
My math teacher told us that it's almost always easier to leave x alone and put the other factor into the denominator, which apparently is not true for the first problem. I guess I need to apply the chain rule to differentiate the denominator, and I didn't do that? Putting the x in the denominator looks easier anyway.
The back-of-the-book answer to the second problem is e^-2. I'm pretty sure I'm supposed to use ln to bring the exponent down; unfortunately I'm just not sure what to do with it once I have it as ln(1-2x)/x.
edit: Savant, you're right. The second problem is lim(x->ZERO), not infinity, which simplifies things a bit.
Yeah, that makes it a lot more straightforward.
So the method you want to use for that is the one I mentioned earlier, where you do the following transformation:
lim(x->0) (1-2x)^(1/x) = lim(x->0) e^( ln((1-2x)^(1/x)) ) = e^( lim(x->0) (1/x)ln(1-2x) )
Be careful about moving around limits like that into functions too much in general, but you can get away with it for the exponential.
Anyways, you can use l'Hospital's rule again to find lim(x->0) ln(1-2x)/x, which is a pretty straightforward application of it. Then plug that back in to the above equation.
Well, l'Hôpital if you want to get super specific.
My textbook has it down as l'Hospital, and that's how my professor wrote it. Wikipedia agrees with you, though.