2
$\begingroup$

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 $

enter image description here

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

enter image description here

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.