Evaluate $\lim_{h\to 0}\frac{1}{h^3e^{\frac{1}{\sqrt{h^2}}}}$

Question

Background: I need to find if \begin{cases} \frac{1}{x^2+y^2+z^2}e^{-\frac{1}{\sqrt{x^2+y^2+z^2}}}, \text{ }(x,y,z)\neq (0,0,0)\\ 0,\text{ } (x,y,z)=(0,0,0) \end{cases}

Differentiable at $(0,0)$

So I found that the function is continues at (0,0,0) and now I am checking if there are partial derivatives

a. why we must derive by definition?

So I derived by definition and got

$$\lim_{h\to 0}\frac{1}{h^3e^{\frac{1}{\sqrt{h^2}}}}$$

I would use substitution to get to an expression of the form $\frac{t^3}{e^t}$ but I have even and odd powers

Answer 1

\begin{align*} \lim_{h\rightarrow 0^{+}}\dfrac{1}{h^{3}e^{1/h}}&=\lim_{u\rightarrow\infty}\dfrac{u^{3}}{e^{u}}\\ &=\lim_{u\rightarrow\infty}\dfrac{3u^{2}}{e^{u}}\\ &=\lim_{u\rightarrow\infty}\dfrac{6u}{e^{u}}\\ &=\lim_{u\rightarrow\infty}\dfrac{6}{u}\\ &=0, \end{align*} \begin{align*} \lim_{h\rightarrow 0^{-}}\dfrac{1}{h^{3}e^{-1/h}}&=\lim_{u\rightarrow\infty}-\dfrac{u^{3}}{e^{u}}\\ &=-\lim_{u\rightarrow\infty}\dfrac{u^{3}}{e^{u}}\\ &=0. \end{align*}

Answer 2

You need to differentiate by definition, because you have no formula for the function at zero and including neighbourhoods.

For your limit, $$ \lim_{h\to0^+}\frac{1}{h^3e^{1/|h|}} =\lim_{x\to\infty}\frac{x^3}{e^{x}}=0, $$ easily done by L'Hôpital. Similarly, $$\lim_{h\to0^-}\frac{1}{h^3e^{1/|h|}}=0.$$

Answer 3

The story for Differentiability and Continuity for functions of several variables is as follow:

1. If a function is discontinuos then it can't be differentiable, infact continuity is a necessary condition since differentiability implies continuity.

2. If partial derivatives do not exist then $f$ can't be differentiable, infact existence of partial derivatives is a necessary condition since differentiability implies their existence.

3. If partial derivatives exist and are continuos you are done, infact for the "Differentiability theorem" if all the partial derivatives exist and are continuous in a neighborhood of the point then (i.e. sufficient condition) the function is differentiable at that point.

4. If partial derivatives are not continuos at the point you can't yet conclude anything about differentiability. You need to check directly differentiability by definition that:

$$\lim_{h\rightarrow 0} \frac{\| f(h)-f(0)-\Delta f(0)\cdot h\|}{\|h\|}=0$$

In this case by the limit you can easy verify that partial derivatives are $0$ at $(x,y,z)=(0,0,0)$ and you can conclude by definition according to point 4 .

Answer 4

Separating the cases for positive and negative $h$, we get rid of the square root:

$$ \lim_{h\to 0^+}{\frac{1}{h^3 e^{\frac{1}{\sqrt{h^2}}}}}=\lim_{h\to 0^+}{e^{-\ln(h^3)- \frac{1}{h}}} $$

$$ \lim_{h\to 0^-}{\frac{1}{h^3 e^{\frac{1}{\sqrt{h^2}}}}}=\lim_{h\to 0^-}{-e^{-\ln(-h^3)+ \frac{1}{h}}} $$

Evaluating:

$$ \lim_{h\to 0^+} \left( {-\ln(h^3)- \frac{1}{h}} \right) $$

And: $$ \lim_{h\to 0^-} \left( {-\ln(-h^3)+ \frac{1}{h}} \right) $$ Note that:

$$ \lim_{x\to 0} x\ln(x) = \lim_{x\to 0} \frac{\ln(x)}{1/x}=\lim_{x\to 0} \frac{1/x}{-1/x^2}=\lim_{x\to 0} (-x)=0 $$

Then: $$ \lim_{h\to 0^+} \left( {-\ln(h^3)- \frac{1}{h}} \right) = \lim_{h\to 0} -\frac{3h\ln (h)+1}{h}= -\infty $$

And $$ \lim_{h\to 0^-} \left( {-\ln(-h^3)+ \frac{1}{h}} \right) = \lim_{h\to 0} -\frac{3h\ln (h)-1}{h} = -\infty $$

In both cases the original limit is of the form $e^{-\infty}=0$ because the inverse square root term dominates over the logaritmic term.