- #1

Vale132

- 7

- 0

## Homework Statement

Let

f(x) =\begin{cases}

0 & \text{ if } x\leq 0 \\

e^\left ( -1/x \right ) & \text{ if } x> 0

\end{cases}

Compute f'(x) for x < 0 and x > 0.

## Homework Equations

[tex] f'(x) = \lim \ \ \ \ \ \ \displaystyle{\frac{e^{(-1/(x+h)} - e^{-1/x}}{h}} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ h\rightarrow0[/tex]

## The Attempt at a Solution

Did the x < 0 part already, seemed straightforward (got 0/h = 0).

For x > 0:

[tex]f'(x) = \lim \ \ \ \ \ \ \frac{1}{he^{1/(x+h)}} - \ \ \ \ \lim \ \ \ \ \ \frac{1}{he^{1/x}} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ h\rightarrow0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ h\rightarrow0[/tex]

It is it true that because, as h -> 0 the h in the exponential's denominator (in the first limit) goes to zero, the two limits are approximately the same (large) quantity, so their difference is zero?

**EDIT**: Oops, looks like I broke up the limit incorrectly, since it's not a limit of two separate functions of x. Should I instead take the limit of the difference of two quotients?

Last edited: