[SOLVED] Find values of a, b and c such that limit ... is finite?

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January 16th 2009, 07:40 PM
fardeen_gen

[SOLVED] Find values of a, b and c such that limit ... is finite?

Find values of a, b and c such that:
lim (cos 4x + a cos 2x + b)/x^4 = Finite
x --> 0
January 16th 2009, 08:46 PM
Isomorphism

Quote:

Originally Posted by fardeen_gen

Find values of a, b and c such that:
lim (cos 4x + a cos 2x + b)/x^4 = Finite
x --> 0

You have missed a 'c' somewhere, thats why we will not be getting any answer correctly. But I will tell you the general idea...

Let L be the limit.

Observe that if $g(0) = 0$ and $\lim_{x \to 0} \frac{f(x)}{g(x)} =$ L, then $\lim_{x \to 0} f(x) = 0$ .

Applying it here, we get $1 + a + b = 0$ .

Now apply L'Hospital's rule, to get another form for L. Do the same process again.

Alternate trick is to substitute the power series for cos and choosing coefficients such that the limit exists.
January 16th 2009, 08:55 PM
fardeen_gen

No missing 'c' according to text.(Worried)
January 16th 2009, 09:11 PM
Isomorphism

Quote:

Originally Posted by fardeen_gen

Find values of a, b and c such that:
lim (cos 4x + a cos 2x + b)/x^4 = Finite
x --> 0

Quote:

Originally Posted by fardeen_gen

No missing 'c' according to text.(Worried)

Then why does the question say find a, b and c?
January 16th 2009, 09:21 PM
Isomorphism

You can however use power series or L'Hospitals to get the following equations:

a+b+1 = 0 and 8 + 2a = 0 and thus a = -4 and b = 3. So the limit L = 8.

To do this using power series, write $\cos t = 1 - \frac{t^2}{2!} + \frac{t^4}{4!} + (t^6$ ke terms $....)$ . Then group terms with same powers in the numerator. All terms with constant and power of x^2 must go to 0. That will give you the above two equations....
January 16th 2009, 09:26 PM
fardeen_gen

The question was a general one, under which the specific problem appears(there are some which involve a, b and c, a and b, b and c... and so on).

Thanks for the help!!
I got the same answer right now!