Limit as x approaches 0 for given function

[needingtoknow]

Homework Statement

lim x --> 0 for the function sqrt((1/x^2)-(1/x)) - sqrt((1/x^2)-(1/x))

Analyze the cases x > 0 and x < 0

The Attempt at a Solution

The solutions book simplifies the expression to

2x/(abs(x)*(sqrt(1+x)+sqrt(1-x)))

I know how to evaluate the limit from here. But I'm wondering why did they only place the absolute value sign for the single x in the denominator that multiplies with this (sqrt(1+x)+sqrt(1-x)) expression. Why isn't the absolute value sign around every single x?The answer is that the limit doesn't exist.

 

[gopher_p]

Homework Statement

lim x --> 0 for the function sqrt((1/x^2)-(1/x)) - sqrt((1/x^2)-(1/x))

Analyze the cases x > 0 and x < 0

The Attempt at a Solution

The solutions book simplifies the expression to

2x/(abs(x)*(sqrt(1+x)+sqrt(1-x)))

I know how to evaluate the limit from here. But I'm wondering why did they only place the absolute value sign for the single x in the denominator that multiplies with this (sqrt(1+x)+sqrt(1-x)) expression. Why isn't the absolute value sign around every single x?


The answer is that the limit doesn't exist.

It looks like maybe your problem statement has a typo, but that doesn't seem to matter to the question that you asked.

The absolute value comes into play because $\sqrt{x^2}=|x|$.

 

[mfb]

Why isn't the absolute value sign around every single x?

Why do you expect one?

 

[needingtoknow]

Ohh right I forgot out that equality. So you're saying all they did was sub in abs(x) wherever there as x^2?

 

[Ray Vickson]

Homework Statement

lim x --> 0 for the function sqrt((1/x^2)-(1/x)) - sqrt((1/x^2)-(1/x))

Analyze the cases x > 0 and x < 0

The Attempt at a Solution

The solutions book simplifies the expression to

2x/(abs(x)*(sqrt(1+x)+sqrt(1-x)))

I know how to evaluate the limit from here. But I'm wondering why did they only place the absolute value sign for the single x in the denominator that multiplies with this (sqrt(1+x)+sqrt(1-x)) expression. Why isn't the absolute value sign around every single x?


The answer is that the limit doesn't exist.

Your question must have a "typo" in it; what you have written is $\sqrt{f(x)} - \sqrt{f(x)}$, which is identically 0 for all positive values of $f(x) = 1/x^2 - 1/x.$

 

[mfb]

Ohh right I forgot out that equality. So you're saying all they did was sub in abs(x) wherever there as x^2?

It just appears once. They replaced $\sqrt{x^2}$ by abs(x) (note the square root), because $\sqrt{x^2} = |x|$.

 

[gopher_p]

Ohh right I forgot out that equality. So you're saying all they did was sub in abs(x) wherever there as x^2?

No. I'm saying that they used the equality, $\sqrt{x^2}=|x|$, as part of rewriting the given algebraic expression in way that is more easily analyzed in the context of this problem.

One could say that they replaced certain instances of $\sqrt{x^2}$ with instances of $|x|$.

I don't like the word "substitute" to describe what is going on here. I can't exactly put my finger on why I don't like it. It just seems like the wrong word to use.