- #1

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1) lim x->0 sin2x/sin5x

2) write an equation for the stright line through (1,0) that is tangent to the graph of y= x + 1/x....

thanks in advance

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- Thread starter MercuryRising
- Start date

- #1

- 28

- 0

1) lim x->0 sin2x/sin5x

2) write an equation for the stright line through (1,0) that is tangent to the graph of y= x + 1/x....

thanks in advance

- #2

HallsofIvy

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1) What if this were [itex]lim_{y,z->0}\frac{sin y}{y}\frac{z}{sin z}[/itex]? Could you do it then? Can you think of a way to change it into that form?

2) One definition of "derivative" is that it

- #3

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1) i tried (2x/2x) (sin2x)/sin(5x) -> (2x/sin5x) (sin2x)(2x)..now i dont know how to get rid of the (2x/sin5x)

2) hmm...the derivative i got was (X^2-1)/x^2. do i just plug in a random number to get a slope then use point sliope formula to make it include (1,0)?

- #4

Diane_

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2) You don't use a random point. You have to find a point (x, y) such that the slope of the line (which is constant, of course) equals the slope of the curve at that point. Note that the line will also pass through that point. Think simultaneous equations.

- #5

HallsofIvy

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MercuryRising: You changed [tex]\frac{sin 2x}{sin 5x}[/tex] to [tex]\frac{sin 2x}{2x}\frac{5x}{sin 5x}[/tex] by multiplying the numerator by 5x and denominator by 2x. Of course you

- #6

Diane_

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HallsofIvy said:L'Hopital's rule will work (easily) but I would consider it overkill! You wouldn't want to do this theeasyway, would you?

As a matter of fact, yes.

- #7

Tom Mattson

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Diane_ said:Does this suggest anybody's Rule?

Maybe

- #8

HallsofIvy

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Do you have a cold, Tom?

- #9

Tom Mattson

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No, I do not habe a cold. :tongue2:

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