Solving for the Limit: lim x->0 (3/x^2 - 2/x^4)

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The limit lim x->0 (3/x^2 - 2/x^4) is evaluated by combining the fractions into a single expression. Simplifying leads to (4x^2 - 2)/x^4, where the numerator approaches -2 and the denominator approaches 0 as x approaches 0. This results in the limit tending towards -infinity. Although some participants express confusion about undefined operations, the consensus is that the limit does not exist in the traditional sense, as it diverges to negative infinity. The discussion emphasizes the importance of proper fraction manipulation to arrive at the conclusion.
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Homework Statement


Find the limit: lim x->0 (3/x^2 - 2/x^4)


Homework Equations





The Attempt at a Solution


I've tried manipulating it in a lot of ways, but nomatter what I do I've still either got the limit of the denominator = 0 or one of the terms in either the num. or den. being divided by x (which is tending to zero)

I've got as far as lim x->0 ( 9x^4 - 4 / 3x^6 + 2x^4 ) but this doesn't help as the den. = 0.

I've graphed this function and had a look - the limit should equal -(infinity).

Utterly confused.
 
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Welcome to PF caesius.

Graphing the function was a good idea, at least now you know the answer :smile:

You have
\lim_{x\to0} \left( \frac{3}{x^2} - \frac{2}{x^4} \right)
Standard approach in these cases: write it as a single fraction. Though in principle you did this correctly, you could have done it much more easily by just multiplying the first term only by x^2 / x^2 and then adding the fractions. Of what you then get it is much easier to show that it tends to -infinity.
 
Ok, I get (3x^2 - 2) / x^4 but I don't see how this helps.

If x was tending to infinity I see how to solve it but for 0?
 
Divide numerator and denominator by x^4.
 
I've done that but I get 3x^-2 - 2x^-2 / 1

Which to me looks like: (undefined) - (undefined) / 1

How can that have a solution?
 
Isn't "the limit does not exist" a solution?
 
Ok I've just read how to format properly, for clarity I have:

\frac{\frac{3}{x^2} - \frac{2}{x^2}}{1}

which when taking the limit as x tends to zero will result in indefined operations no?
 
HallsofIvy said:
Isn't "the limit does not exist" a solution?

But I know the "limit" is -\infty, and I'm aware that this is not a limit by definition, surely I should be able to show this from the equation.
 
Okay, here's what I would have done: combine the two fractions to get
\frac{3}{x^2}- \frac{2}{x^4}= \frac{4x^2- 2}{x^4}
Since the numerator goes to -2 while the denominator goes to 0, the fraction itself goes to -\infty (i.e. the limit does not exist). What more do you want?
 
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