Limit help(1/rad(x) - 1/rad(x^2 + x))

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In summary, the given expression is evaluated by manipulating it and applying L'Hopital's rule to get a limit of 0/1, indicating that the limit as x approaches 0+ is equal to 0.
  • #1
dlp211
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Homework Statement



lim x->0+ 1/x1/2 - 1/(x2+x)1/2

Homework Equations



inf - inf

The Attempt at a Solution



I can't get x out of the bottom of the equation.
 
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  • #2
I have manipulated the equation to:

x1/2(x+1) - (x2+x)1/2 / x(x+1)
 
  • #3
Which I now was able to reduce to:

x(x1/2-x) / x(x+1)

which goes to:

(x1/2-x) / (x+1)

and so my limit is 0/1 which is zero.

Anyone want to check the work I would greatly appreciate it.
 
  • #4
Yeah, that limit is tending to 0. It's important that it's from the positive side because of the domain of the radicals.

I simply combined the fractions and the limit went to the indeterminate form 0/0. Then I just L'Hopital ruled it into 0/1
 
Last edited:
  • #5
dlp211 said:
I have manipulated the equation to:

x1/2(x+1) - (x2+x)1/2 / x(x+1)
This is not an equation - it's an expression. An equation has two expressions that are connect by =.
 

1. What is the limit of the function (1/rad(x) - 1/rad(x^2 + x)) as x approaches infinity?

The limit of the function as x approaches infinity is equal to 0.

2. Can the limit of this function be evaluated using L'Hopital's rule?

Yes, L'Hopital's rule can be applied to this function to evaluate the limit.

3. Is the limit of this function defined at x=0?

No, the function is undefined at x=0 as it would result in division by 0.

4. How does the value of the limit change as x approaches 0 from the left and right?

The limit from the left side of x=0 is -1/2, while the limit from the right side is 0. Therefore, the limit does not exist at x=0.

5. Does the limit of this function have any applications in real-world problems?

Yes, the limit of this function can be used in physics and engineering to calculate the rate of change of a quantity over time.

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