Check if the function has a vertical tangent at origin.

$y= -\sqrt{|x|}\hspace {10pt} for \hspace {10pt} x\leq0 $

$y= \hspace {12pt}\sqrt{x}\hspace {12pt} for \hspace {12pt} x>0 $

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NOTE: Ignore the 4th quadrant red graph

$\lim_{h\to0^+} \frac{f(x+h)-f(x)}{h}=\lim_{h\to0^+}\frac{\sqrt{x+h}-0}{h}=\lim_{h\to0^+}\frac{\sqrt{x+h}}{h}=\lim_{h\to0^+}\frac{\sqrt h}{h}=\lim_{h\to0^+}\frac{1}{h^{1/2}}=+\infty$

Now, for $\lim_{h\to0^-}$ from the graph $f(x) > f(x+h)$

$\therefore \lim_{h\to0^-}\frac{f(x) - f(x-h)}{h}=\lim_{h\to0^-}\frac{0 - (-\sqrt{|x-h|})}{h}=\lim_{h\to0^-}\frac{\sqrt{|x-h|}}{h}=\lim_{h\to0^-}\frac{\sqrt h}{h}=\lim_{h\to0^-}\frac{1}{\sqrt h}$

But $h$ is a very small negative number and $\sqrt{.}$ of a negative number is not deined, so I conclude the vertical tangent is not present at origin.

But according to the solution manual, the second limit is also $\infty$ and hence the vertical tangent is present.

This is how they solve it

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The problem comes in the second to the last equality. Your equality was: $lim_{h→0^-}\frac{\sqrt{|x−h|}}{h}=lim_{h→0^-}\frac{\sqrt{h}}{h}$

However, as $h$ is negative, $|x−h|=-h$ when $x$ is $0$.

Thus the equality should be $lim_{h→0^-}\frac{\sqrt{|x−h|}}{h}=lim_{h→0^-}\frac{\sqrt{-h}}{h}=lim_{h→0^-}\frac{1}{\sqrt{-h}}=+\infty$

Hope that helped.


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