- #1

Qube

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## Homework Statement

http://i.minus.com/j1DgeK5VMMfHn.png [Broken]

## Homework Equations

To simplify a limit with a square root multiply by the conjugate.

## The Attempt at a Solution

The limit doesn't seem to exist as the left and right hand limits diverge.

The left hand limit with f(x) = x^2 + 4 results in x^2/x = x and the limit of x as x approaches 0 from the left is 0.

The right hand limit, after multiplying it by its conjugate, simplifies to 1/[(sqrt(x+16)) + 4]. Taking the limit of this as x approaches 0 results in 1/8.

The limit definition of the derivative for this particular f(x) doesn't seem to exist.

Does this imply the function is not differentiable at x = 0? I graphed it out and it seems there is a corner at x = 0. I tried looking up the mathematical definition of a corner but the best I got was that at a corner the derivative changes abruptly; it's not "smooth." The derivative does jump from being negative on the left with x^2 + 4 and to being positive with the sqrt(x+16). So what's the mathematical definition of a corner? I know for a cusp the mathematical definition is that the left and right hand limits go to infinities of different signs at the point of the cusp. So what's a corner? And am I correct that this function is not differentiable at x = 0?

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