Limit Question: 1/|x| - sqrt[(x + 1)/x^2]

[o_O]

Question:
For \(\displaystyle x \geq -1\), let:
\(\displaystyle f(x) = \frac{1}{|x|} - \sqrt{\frac{x + 1}{x^{2}}}\)
Check if the discontinuity at x = 0 is removable or not.

Attempt:
Using a graphing calculator, I can see that there is a jump discontinuity at x = 0 from y = 0.5 to y = -0.5. However, I'm having trouble getting the limit of f(x) as x approaches zero to equal those y values from both sides:

\(\displaystyle \lim_{x \to 0^{-}} f(x) = \lim_{x \to 0^{-}} \frac{1}{-x} - \sqrt{\frac{x + 1}{x^{2}}\) (x is negative)

For the sake of the texing, let the \(\displaystyle \lim_{x \to 0^{-}}\) be implied.

\(\displaystyle = \frac{1}{-x} - \frac{\sqrt{x+1}}{\sqrt{x^{2}}\)

\(\displaystyle = \frac{1}{-x} - \frac{\sqrt{x+1}}{-|x|}\)

\(\displaystyle = \frac{1}{-x} - \frac{\sqrt{x+1}}{x}\) (x is again negative)

\(\displaystyle = \frac{1}{-x} + \frac{\sqrt{x+1}}{-x}\)

\(\displaystyle = \frac{1 + \sqrt{x+1}}{-x} \left(\frac{1-\sqrt{x+1}}{1-\sqrt{x+1}}\right)\)

\(\displaystyle = \frac{1 - x - 1}{-x\left(1-\sqrt{x+1}\right)}\)

\(\displaystyle = \frac{-x}{-x\left(1-\sqrt{x+1}\right)}\)

\(\displaystyle = \frac{1}{\left(1-\sqrt{x+1}\right)}\)

\(\displaystyle = \frac{1}{\left(1-\sqrt{0+1}\right)}\)

\(\displaystyle = \frac{1}{0}\)

What did I do wrong? It should be \(\displaystyle \frac{1}{2}\) ...

 

[Deleted member 4993]

When x is negative

\(\displaystyle sqrt(x^2) = -n\) (when x = -n and n is positive) - try to work this way

 

[o_O]

\(\displaystyle = \frac{1}{-x} - \frac{\sqrt{x+1}}{\sqrt{x^{2}}\)

\(\displaystyle = \frac{1}{-x} - \frac{\sqrt{x+1}}{-n}\)

\(\displaystyle = \frac{1}{n} - \frac{\sqrt{x+1}}{-n}\)

\(\displaystyle = \frac{1}{n} + \frac{\sqrt{x+1}}{n}\)

\(\displaystyle = \frac{1 - x - 1}{n\left(1 - \sqrt{x+1}\right)\)

Continuing on ... this will eventually wind up with a 0 denominator wouldn't it?

 

[Deleted member 4993]

However

|x| = n

 

[o_O]

But that would just be the same thing as what I started. I don't see how replacing |x| in my method with n would help out.

 

[stapel]

I don't think there's any need to replace "x" with "n". The definition of absolute value is that, for x < 0, the absolute value of x, being sqrt[x²], is -x, where here the "minus" means "the opposite sign of", not "negative".

For instance, if x = -3 < 0, then |x| = sqrt[(-3)²] = sqrt[9] = 3 = -(-3) = -x.

For x < 0, you then have:

. . . . .1/|x| - sqrt[(x + 1) / x²]

. . . . .1/(-x) - sqrt[x + 1] / sqrt[x²]

. . . . .-1/x - sqrt[x + 1] / (-x)

. . . . .-1/x + sqrt[x + 1] / x

. . . . .(sqrt[x + 1] - 1) / x

Then do the "multiplication by the conjugate" thing, which leads to:

. . . . .[(x + 1) - 1] / [x (sqrt[x + 1] + 1)]

. . . . .[x] / [x (sqrt[x + 1] + 1)]

. . . . .1 / (sqrt[x + 1] + 1)

As x tends toward zero, you get 1/(1 + 1) = 1/2, or 0.5.

When you take the limit from the right, then |x| = x, and sqrt[x²] = x. So the only difference is the sign on the denominator. But this sign change, naturally, leads to a limit of -0.5 instead of +0.5.

Hope that makes sense!

Eliz.

 

[Deleted member 4993]

Stapel,

You are absolutely right - no need for 'n' substitution. I did not spend enough time thinking about it.

 

[o_O]

Oooh. That makes a bit more sense. So when we talk about \(\displaystyle \sqrt{x^{2}\), we are only referring the positive value of x? Thanks for the help guys!