- #1

- 97

- 0

cos(x) - 1 / sin(x)

I can't really get anything going, I have:

cos(x)/sin(x) - 1/sin(x)

=cot(x) - csc(x)

but that still leaves the limit undefined.

any help?

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- Thread starter nothing123
- Start date

- #1

- 97

- 0

cos(x) - 1 / sin(x)

I can't really get anything going, I have:

cos(x)/sin(x) - 1/sin(x)

=cot(x) - csc(x)

but that still leaves the limit undefined.

any help?

- #2

- 1,235

- 1

Use L'Hopitals rule.

- #3

radou

Homework Helper

- 3,120

- 6

- #4

Office_Shredder

Staff Emeritus

Science Advisor

Gold Member

- 4,636

- 648

Use parentheses.

cos(x) - 1 / sin(x)

becomes (cos(x) - 1) / sin(x) and it makes a lot more sense. L'Hopital is overkill here, as long as you know the limit of sin(x)/x = 1 as x approaches zero.

So (cos(x) - 1)/sin(x) = x(cos(x) - 1)/(xsin(x). But x/sin(x) approaches 1, so we'll factor that out and leave ourselves with (cos(x) - 1)/x as x approaches zero. You should recognize this as the derivative of cos(t) at t=0

As a quick point, you shouldn't actually have instantly realized that equals the derivative (I didn't either until someone pointed it out to me a week ago that you can solve problems this way :) ), but it's nice to know. I would have left you there to figure it out, but it's really not in the most recognizable form

cos(x) - 1 / sin(x)

becomes (cos(x) - 1) / sin(x) and it makes a lot more sense. L'Hopital is overkill here, as long as you know the limit of sin(x)/x = 1 as x approaches zero.

So (cos(x) - 1)/sin(x) = x(cos(x) - 1)/(xsin(x). But x/sin(x) approaches 1, so we'll factor that out and leave ourselves with (cos(x) - 1)/x as x approaches zero. You should recognize this as the derivative of cos(t) at t=0

As a quick point, you shouldn't actually have instantly realized that equals the derivative (I didn't either until someone pointed it out to me a week ago that you can solve problems this way :) ), but it's nice to know. I would have left you there to figure it out, but it's really not in the most recognizable form

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- #5

- 145

- 0

[cos(0)-1]/sin(0) is the indeterminate for 0/0.

L'Hopitals rule says that if your limit is in such an intermanite form, the limits of the derivatives of the top and bottom will be the same... I.E

lim f(x) f'(x)

x-->N ------ = -------

g(x) g'(x)

2 Notes... This only works if the lim is an indeterminate for (inf./inf., 0/0, 1^inf. etc.

and that its NOT [f(x)/g(x)]' its the derivitive of the top alone and the derivitave of the bottom alone

In this example the derivative of the top is (-sin(x))

The deriviative of the bottom is cos(x)

so the new limit becomes lim -sin(x)

x-->0 --------- = 0

cos(x)

The final answer is 0. Hope that helps

- #6

- 145

- 0

kinda messed up the margins, you can fingure it out

- #7

- 97

- 0

btw, the lim x-> 0 of x/sin(x) is 1?

- #8

- 11

- 0

nothing123 said:btw, the lim x-> 0 of x/sin(x) is 1?

Yes, it is. Similarly, [tex]\lim_{x\rightarrow 0} \frac{sin(x)}{x} = 1[/tex]. The intuitive (read: non-Calc) way to see this is to set up a table with columns x and sin(x), and use your calculator to see what happens to sin(x) as your x value gets closer to (but not equal to) 0. You should see that as x approaches 0 in your table, sin(x)-x also approaches 0. The calculus-based way of seeing this is to use L'Hospital's Rule, since [tex] \frac{sin(x)}{x} = \frac{0}{0} [/tex](indeterminate form) when x=0. The derivative of sin(x) is with respect to x is cos(x), and the derivative of x is 1. Then, [tex]\lim_{x\rightarrow 0} cos(x) = 1[/tex].

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