Deeper understanding of tangent line and Limit I am trying to understand limits, but I constantly manage to get the answers wrong. This is what I am struggling with: $$ f(x)= \left\\{ \begin{array}{cc} \sqrt{x} & x \ge 0 \\\ -\sqrt{-x} & x < 0 \\\ \end{array} \right. $$ The question is, does this function have a limit at $x=0$ and what is the tangent line? What I do is substituting the value $x=0$ into the equation: $$ \lim_{h\to 0} \, \frac{\sqrt{h+x}-\sqrt{-x}}{h} $$ This gives me: $$ \frac{1}{\sqrt{h}} $$ Now as $h$ approaches zero we end up with an infinite large number. How is it that the tangent line to this is $0$?

Here is a diagram of the situation: !enter image description here

The limit as $x\to0$ exists, because the left-handed and the right-handed limit approach the same value.

According to the diagram, it seems that the tangent line is vertical. When the tangent line is vertical the slope is infinite.

Let's see if the slope is indeed infinite:

$$ f'(x)= \left\\{ \begin{array}{cc} \frac{1}{2\sqrt{x}} & x \ge 0 \\\ \frac{1}{2\sqrt{-x}} & x < 0 \\\ \end{array} \right. $$

For $x=0$ the slope approaches infinity, so the tangent line is vertical.