Limit as x approaches 0 for given function

In summary, the given expression can be simplified to 2x/(abs(x)*(sqrt(1+x)+sqrt(1-x))), and the absolute value sign is only placed around the x in the denominator because of the equality sqrt(x^2) = |x|. This simplification makes it easier to analyze the limit, and the reason for using the absolute value sign is to account for the different cases of x > 0 and x < 0.
  • #1
needingtoknow
160
0

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.
 
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  • #2
needingtoknow said:

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|##.
 
  • #3
Why isn't the absolute value sign around every single x?
Why do you expect one?
 
  • #4
Ohh right I forgot out that equality. So you're saying all they did was sub in abs(x) wherever there as x^2?
 
  • #5
needingtoknow said:

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.##
 
  • #6
needingtoknow said:
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|##.
 
  • #7
needingtoknow said:
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.
 

Related to Limit as x approaches 0 for given function

What is the definition of a limit as x approaches 0?

The limit as x approaches 0 for a given function is the value that the function approaches as the input (x) gets closer and closer to 0, without actually reaching 0.

How is the limit as x approaches 0 calculated?

The limit as x approaches 0 is calculated by evaluating the function at values of x that are closer and closer to 0, both from the left and the right side. If the function approaches the same value from both sides, that value is the limit as x approaches 0. If the function approaches different values from the left and the right side, then the limit does not exist.

Why is the limit as x approaches 0 important?

The limit as x approaches 0 is important because it helps us understand the behavior of a function near a specific point. It can also help us determine if a function is continuous at that point, which is an important concept in calculus and real analysis.

Can the limit as x approaches 0 be infinite?

Yes, the limit as x approaches 0 can be infinite. This occurs when the function approaches positive or negative infinity as x gets closer and closer to 0.

What are some common techniques for evaluating the limit as x approaches 0?

Some common techniques for evaluating the limit as x approaches 0 include factoring, simplifying, and using algebraic manipulation. Additionally, L'Hopital's rule and the squeeze theorem can also be helpful in certain cases.

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