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I've tried evaluating, but only end up with 2xsin(1/x), and that is incorrect. I have also not learned L'Hospital's rule yet so any help would be great!

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- Thread starter MathewsMD
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I've tried evaluating, but only end up with 2xsin(1/x), and that is incorrect. I have also not learned L'Hospital's rule yet so any help would be great!

- #2

Office_Shredder

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Assuming the latter, you need to use the product rule

- #3

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Assuming the latter, you need to use the product rule

It is the latter. And you can only evaluate using limits.

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Pls post your attempt.

- #5

Office_Shredder

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=lim [[(x+h)^2]sin(1/x+h) - (x^2)sin(1/x)]/h

h→0

=lim [(x^2)sin(1/x+h) + 2xhsin(1/x+h) + (h^2)sin(1/x+h) - (x^2)sin(1/x)]/h

h→0

=[lim[(x^2)sin(1/x+h) - (x^2)sin(1/x)] + lim[2xhsin(1/x+h) + (h^2)sin(1/x+h)]]/lim h

h→0 h→0 h→0

= 0 + lim [[h (2xsin(1/x+h) + hsin(1/x+h)]/h]

h→0

= lim (2xsin(1/x+h) + hsin(1/x+h)]

h→0

= 2xsin(1/x)

If there are any errors in my steps, please point them out. This is not the correct answer (I believe), and if someone could show me the correct way to evaluate this limit, both when x = 0 and when x ≠ 0, that would be great. Thanks!

EDIT: The h→0 should be everywhere you see the lim notation. Sorry, I can't really format well here.

- #7

Office_Shredder

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[tex] \lim_{h\to \0} \frac{ x^2 \sin(1/(x+h)) - x^2 \sin(1/x)}{h} = 0 [/tex]

which is not correct.

- #8

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I meant:

[tex] \lim_{h\to\ 0} {x^2 \sin(1/(x+h)) - x^2 \sin(1/x)} = 0 [/tex]

Then I assigned the limit to the denominator (h) and since the numerator was just addition, I did the same for the last two terms.

[tex] \lim_{h\to\ 0} {x^2 \sin(1/(x+h)) - x^2 \sin(1/x)} = 0 [/tex]

Then I assigned the limit to the denominator (h) and since the numerator was just addition, I did the same for the last two terms.

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- #9

Office_Shredder

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I see what you are doing now - it was hard to read in your last post but you cannot move around limits and fractions like that. Once you get to this point

[tex] \frac{ x^2 \sin(1/(x+h)) - x^2 \sin(1/x) + 2xh\sin(1/(x+h)) + h^2 \sin(1/(x+h))}{h} [/tex]

You cannot re-write this as

[tex] x^2 \sin(1/(x+h)) - x^2 \sin(1/x) + \frac{2xh \sin(1/(x+h)) + h^2 \sin(1/(x+h)}{h} [/tex]

[tex] \frac{ x^2 \sin(1/(x+h)) - x^2 \sin(1/x) + 2xh\sin(1/(x+h)) + h^2 \sin(1/(x+h))}{h} [/tex]

You cannot re-write this as

[tex] x^2 \sin(1/(x+h)) - x^2 \sin(1/x) + \frac{2xh \sin(1/(x+h)) + h^2 \sin(1/(x+h)}{h} [/tex]

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[tex][ \lim_{h\to\ 0}{(x^2 \sin(1/(x+h)) - x^2 \sin(1/x)) + \lim_{h\to\ 0}(2xh\sin(1/(x+h)) + h^2 \sin(1/(x+h))}] ÷ \lim_{h\to\ 0}{h} [/tex]

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Okay. So what method would be best in this case to overcome this problem?

I don't think factoring is really an option here and I don't see a good term to use to rationalize.

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[tex]\lim_{h\to 0} \frac{ x^2\sin(1/(x+h)) - x^2 \sin(1/x)}{h} + \lim_{h\to 0}\frac{ 2xh \sin(1/x) + h^2 \sin(1/x)}{h} [/tex]

- #15

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[tex]\lim_{h\to 0} \frac{ x^2\sin(1/(x+h)) - x^2 \sin(1/x)}{h} + \lim_{h\to 0}\frac{ 2xh \sin(1/x) + h^2 \sin(1/x)}{h} [/tex]

The right side would simplify to 2xsin(1/x) but it's the left side that I'm having trouble with. You can factor out the x^2 and then have the definition of the derivative of sin(1/x) which equals cos(1/x).

I know there are proofs to go from the left to the right side, but I was trying to understand and devise my own method I guess, which was the harder part.

Thanks for the help.

- #16

Office_Shredder

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You can factor out the x^2 and then have the definition of the derivative of sin(1/x) which equals cos(1/x).

This is not quite right.

I know there are proofs to go from the left to the right side, but I was trying to understand and devise my own method I guess, which was the harder part.

I don't understand what you mean by going from the left to the right side... the left to right side of what I wrote? Those two halfs of the expression are fairly independent.

- #17

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This is not quite right.

I don't understand what you mean by going from the left to the right side... the left to right side of what I wrote? Those two halfs of the expression are fairly independent.

Oh sorry. I initially wrote [tex]\lim_{h\to 0} \frac{\sin(1/(x+h)) - \sin(1/x)}{h} [/tex]

Took it out since I'm still learning to format and it wasn't coming out properly. Sorry for the poor formatting here.

And can't you factor the x^2 out of this limit?

[tex]\lim_{h\to 0} \frac{ x^2\sin(1/(x+h)) - x^2 \sin(1/x)}{h} [/tex]

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