Limit of ((x2+y2+1)1/2) - 1: Evaluate & Simplify

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In summary, the limit of ((x2+y2+1)1/2) - 1 as x and y approach 0 is evaluated by converting to polar coordinates and using the fact that x^2+y^2=r^2. This gives a different result than simplifying the denominator and plugging in 0, which is confirmed by W|A. It is suggested to use polar coordinates instead and to potentially apply L'Hôpital's Rule.
  • #1
Jamie2
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I got that the limit equals 0 by simplifying the denominator from:

((x2+y2+1)1/2) - 1

to
((x2 - (y+1)(y-1))1/2) - 1

then
((x2 - (y(1+1)(1-1))1/2) - 1

and then evaluating the limit by plugging in 0, getting 0/-1=0

is this correct? is there a better way to do it?
 

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  • #2
I get a different result by converting to polar. This result is confirmed by W|A. So, I suggest using polar coordinates...what do you find?
 
  • #3
Could you explain in more detail? I don't think I understand what you mean
 
  • #4
Essentially, I used $x^2+y^2=r^2$ and the limit becomes:

\(\displaystyle \lim_{r\to0}\left(\frac{r^2}{\sqrt{r^2+1}-1} \right)\)

And then I rationalized the denominator. You could also at this point use L'Hôpital's Rule.
 

FAQ: Limit of ((x2+y2+1)1/2) - 1: Evaluate & Simplify

What is the limit of ((x2+y2+1)1/2) - 1 as x and y approach 0?

The limit of ((x2+y2+1)1/2) - 1 as x and y approach 0 is 0. This can be found by substituting 0 for x and y in the expression and simplifying.

Can the limit of ((x2+y2+1)1/2) - 1 be evaluated using L'Hôpital's rule?

No, L'Hôpital's rule can only be used for limits involving quotients of functions, not for expressions with multiple variables like this one.

How can the limit of ((x2+y2+1)1/2) - 1 be simplified?

The expression can be simplified by factoring out a common factor of (x2+y2+1)1/2, leaving the limit as 1 - 1, which simplifies to 0.

Is it possible to find the limit of ((x2+y2+1)1/2) - 1 at a specific point?

No, the limit of a function can only be evaluated as x and y approach a certain value, not at a specific point.

How does the limit of ((x2+y2+1)1/2) - 1 relate to the graph of the function?

The limit of the expression is equivalent to the y-intercept of the graph of the function. In other words, as x and y approach 0, the function approaches the y-axis at a value of -1.

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