Understanding Limits and Derivatives: Solving for x in Trig Functions

[nothing123]

Find the limit as x-> 0 of

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?

 

[courtrigrad]

Use L'Hopitals rule.

 

[radou]

Use the fact that $$\lim_{x\rightarrow a}(F(x) - G(x))=\lim_{x\rightarrow a}F(x) - \lim_{x\rightarrow a}G(x)$$.

 

[Office_Shredder]

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

 

[SNOOTCHIEBOOCHEE]

Using l'hospital's rule is the correct solution...

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

l'hospital's 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 derivative 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

 

[SNOOTCHIEBOOCHEE]

kinda messed up the margins, you can fingure it out

 

[nothing123]

thanks guys, i think office_shredder's post was the most appropriate because this question was assigned on the basis that we have no knowledge of l'hospital's rule.

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

 

[driscoll79]

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

Yes, it is. Similarly, $$\lim_{x\rightarrow 0} \frac{sin(x)}{x} = 1$$. 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 $$\frac{sin(x)}{x} = \frac{0}{0}$$(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, $$\lim_{x\rightarrow 0} cos(x) = 1$$.