# Homework Help: Limit with trig functions

1. Jun 1, 2009

### icosane

1. The problem statement, all variables and given/known data

lim (x*sin(x)) / (2-2*cos(x)
x-> 0

3. The attempt at a solution

I remember the trick to these is to get sinx/x and (1-cosx)/x by multiplying by x/x...

When I multiply the expression by x/x I end up with (x^2 sinx)/(2x(1-cosx))
The sinx/x goes to 1, but the x/(1-cos(x)) goes to zero right? Wouldn't that make the whole expression go to zero, as the separate limits are multiplied together? The book says the answer is one...

2. Jun 1, 2009

### Staff: Mentor

x/(1 - cos(x)) is indeterminate (of the form [0/0]), so you can't just say that it's approaching 0 as x approaches 0.

3. Jun 2, 2009

### bartek2009

You need to apply l'Hopital's rule, that is, to differentiate the numerator and denominator of the expression separately, i.e.:

f/g -> f'/g'

and look at the limit x->0, if it still doesn't work (which is actually the case in this problem) you do it again and check if it gives you a finite value (and you go on until it works).

In this case you first get:

[sin(x) + x*cos(x)]/[2*sin(x)]

which is still nasty as sin(0)=0, then you get:

[2*cos(x) - x*sin(x)]/[2*cos(x)]

which gives you 1 for x=0, you can confirm that by plotting the function (see attached picture).

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4. Jun 2, 2009

### HallsofIvy

The answer is clearly NOT one! Write it as
$$\frac{1}{2} sin(x) \frac{x}{1- cos(x)}$$
$$\frac{x}{1- cos(x)}$$
goes to 1 but the remaining sin(x) goes to 0.

5. Jun 2, 2009

### bartek2009

One can try putting this expression into Mathematica:

Code (Text):
Limit[x*Sin[x]/(2 - 2*Cos[x]), x -> 0]

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• ###### notebook.PNG
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6. Jun 2, 2009

### HallsofIvy

Oh, dear, how stupid of me!
$$\frac{x}{1- cos(x)}$$
goes to 0, not 1 so this is of the form "0/0". The limit is NOT 0 as I asserted!

You need to use L'Hopital's rule repeatedly, as everyone else has been saying.

7. Jun 2, 2009

### Bohrok

$$\lim_{x\rightarrow 0}\frac{x}{1- cos(x)}$$
doesn't go to 0 either, but its reciprocal does.

8. Jun 2, 2009

### Cyosis

If you're not allowed to use l'Hopital, which I never was, Taylor expand the numerator and denominator up to second order.

Last edited: Jun 2, 2009
9. Jun 2, 2009

### Bohrok

You don't have to use L'Hopital's rule or Taylor series. You were on the right track with trying to get fractions with x in the denominator. After factoring out the 2, you get

$$\frac {xsinx} {1 - cosx}$$

multiply by $$\frac {1/x} {1/x}$$ to get $$\frac {sinx} {\frac {1 - cosx} {x}}$$

Since
$$\frac{1 - cosx}{x}$$
leaves us with 0 in the denominator, do something with 1 - cosx to get sinx/x.

10. Jun 2, 2009

### icosane

Thanks guys! I found this problem in a textbook in a chapter before L'Hopital's rule was covered.

Bohrok, I think I can see that if I multiply by (1+cosx) / (1+cosx) That the resulting expression can be manipulated using the trig identity 1 - cos(x)^2 = sin(x)^2 and then everything falls into place.

I just realized that my assumption that (1-cos(x))/x goes to 0 as x goes to 0 is wrong. Does it go to infinity by any chance?

11. Jun 2, 2009