# Homework Help: Indertiminate form

1. Mar 24, 2004

### ACLerok

we starting to learn about l'hospital's rule and i can't figure out this problem. It goes:

Find the Limit for the following indeterminate forms of the type "infinity - infinity"

limit as x approaches 0 of [(1/sin x)-(1/x)]

2. Mar 24, 2004

### Severian596

God it's been too long and my memory of particular derivatives is fading, but if you find a common denominator of $$x sin(x)$$ because:

$$\frac{1}{sin(x)}-\frac{1}{x} = \frac{x-sin(x)}{x \ sin(x)}$$

and we call the fraction

$$\frac{f(x)}{g(x)}$$

can you find the derivatives $$f'(x)$$ and $$g'(x)$$ to find the limit of the original form? You'll have to use the chain rule, right? Like I said it has been too long...hehe

UPDATE:
oops, that product rule, not chain rule!

Last edited: Mar 24, 2004
3. Mar 24, 2004

### Severian596

Okay ACLerok I couldn't help myself and gave it my best shot. See if you get the same thing:

let

$$\frac{x-sin(x)}{x \ sin(x)} = \frac{f(x)}{g(x)}$$

then

$$f'(x) = 1-cos(x)$$

and

$$g'(x) = sin(x) + x \ cos(x)$$

so the format is now

$$\lim_{x \rightarrow 0} \ \frac{1-cos(x)}{sin(x)+x \ cos(x)} = \frac{1-1}{0+(0)(1)}$$

that's indeterminate, too, so take the drivative again to get

$$\lim_{x \rightarrow 0} \ \frac{sin(x)}{2cos(x)-x \ sin(x)}$$

after some algebraic hoopla you get

$$\lim_{x \rightarrow 0} \ \frac{1}{\frac{2cos(x)}{sin(x)}-x} = \frac{1}{\infty - 0} = 0$$

So the limit is zero.

NOTE: I updated this page once to correct errors I made during the computation, starting with the SECOND derivative

Last edited: Mar 24, 2004
4. Mar 25, 2004

### ACLerok

Hi and thanks alot for replying. I understand everything you did but one thing. can you please explain why the limit of 2cos(x)/sin(x) as x approaches 0 equals infinity?

5. Mar 25, 2004

2cos(x)/sin(x) is 2/tan(x). tan(0) is 0, so 2cos(x)/sin(x) is 2/0.

6. Mar 25, 2004

### Severian596

Cookiemonster said it all, but I wanted to add that an easy way to spot this sort of thing is punch sin(0.0000001) in the calc. You get a VERY small number, and if you punched a smaller number than 0.0000001 you'd get an even smaller number for the sin() value. Ad infinitum. If you divide by a fraction you multiply the numerator.

As long as the numerator is a positive value (no matter how big or how small) you get positive infinity for 1/sin(x) as x -> 0

7. Mar 25, 2004

### turin

Make sure your calculator is in radians and you should see a moderately interesting result (which, incidently, supports the limit going to zero in the original difference expression).

8. Mar 25, 2004

### ACLerok

isn't 2/0 undefined?

9. Mar 25, 2004

Yes, but lim{x->0+} 2/x is infinity and lim{x->0-} 2/x is -infinity, so either way it goes to a big number.

Edit: I should point out that tan(x) is continuous in the neighborhood of 0, so that's why it goes to a big number.

Last edited: Mar 26, 2004
10. Mar 26, 2004

### Severian596

Right. ACLerok keep in mind that when referring to limits as x -> 0 people tend to refer to fractions like

2/0

as

2/(an arbitrarily tiny number that's so close to zero the numerator is infinity)

Thanks for this thread, btw! It was fun to work through.

11. Mar 27, 2004

### ACLerok

ok thanks alot guys! one down two more tough ones to go...

12. Mar 28, 2004

### ACLerok

http://www.eden.rutgers.edu/~cjjacob/images/limit.gif [Broken]

is this problem exactly the same as the first one i posted or is there something i have to do first in order to differentiate?

Last edited by a moderator: May 1, 2017
13. Mar 28, 2004

Try putting the two terms over one denominator and simplifying.

14. Mar 29, 2004

### Severian596

Did you figure that last one out AC?

15. Mar 29, 2004

### ACLerok

no i actually just looked at a friend's answer.

this is the second problem i am given.

http://www.eden.rutgers.edu/~cjjacob/images/calcb.gif [Broken]

after simplifying i get lim x->0, (1-x/x^2)

i figure that the lim x->0, (1/x^2), and x^2 being a very small number, is +infinity.

am i right or wrong?

Last edited by a moderator: May 1, 2017
16. Mar 29, 2004

### Severian596

I think you should not have given up so easily. The problem looks much worse than it actually is.

$$lim \ x \rightarrow 0 \ (\frac{1+x}{x}-\frac{1-x}{x})$$

$$= \frac{1+x-(1-x)}{x} = \frac{1+x-1+x}{x} = \frac{2x}{x} = 2$$

Last edited: Mar 29, 2004
17. Mar 29, 2004

Right
I like it