Electronics I problem
Since your question is not clear to me, I'm going out on a limb and assuming that you are trying to find an equation for \$V_{out}\$ in terms of \$V_{in}\$. I don't know why else you would start analyzing the circuit with \$V_{in} = -\infty\$, because it is not useful unless you are thinking about the diodes' regions of operation. I'm going to take the approach of sweeping \$V_{in}\$ to see how the circuit changes.

First assume the forward voltage for both diodes is the same (\$V_{D1} = V_{D2}\$). I'm also going to assume they are ideal diodes. Start by sweeping \$V_{in}\$ from \$-\infty\$ onwards.

Case 1: \$V_{in} = -\infty \$

Since D1 is completely non-conducting, there is no path current can take through it to travel to R2. However, D2 is on because it is forward biased by the negative voltage. Therefore you are left with the circuit below:







simulate this circuit &ndash; Schematic created using CircuitLab

As you can see, \$V_{out} = -(V_{D2} + E)\$. D1 is off and D2 is on. But what happens when you start increasing \$V_{in}\$ more and more? To answer that question, you should ask yourself, at what voltages do D1 and D2 start or stop conducting?

Case 2: \$V_{in} &gt; -\infty \$

So first let's figure out when D2 turns off. As you increase \$V_{in}\$, the voltage across D2 (in the direction of VD2 in the circuit diagram above) will become smaller and smaller until D2 turns off. This happens when the voltage across resistor R2 (call it \$V_{R2}\$) rises above -(\$V_{D2}+E)\$. Noting that \$V_{R2} = \frac{R_2}{R_2 + R_3} V_{in}\$, D2 turns off when \$V_{in} &gt; -\frac{R_2 + R_3}{R_2}(V_{D2}+E)\$.

EDIT: Corrected this section to satisfy boundary conditions. But what about D1? Note that D1 is in parallel with R3. D1 turns on when the voltage across R3 (call it \$V_{R3}\$) exceeds \$V_{D1}\$. Use the voltage divider equation again to get the relationship between \$V_{in}\$ and \$V_{R3}\$. You'll end up with D1 turning on when \$V_{in} &gt; \frac{R_2 + R_3}{R_3} V_{D1}\$. 

Until D1 turns on, however, the circuit looks like a simple voltage divider:







simulate this circuit

Therefore, in the region \$-\frac{R_2 + R_3}{R_2}(V_{D2}+E) &lt; V_{in} &lt; \frac{R_2 + R_3}{R_3} V_{D1} \$, both D1 and D2 are off, and we have the relationship \$ V_{out} = \frac{R_2}{R_2+R_3} V_{in} \$.

Case 3: \$V_{in} &gt; \frac{R_2 + R_3}{R_3} V_{D1} \$

Now, D1 turns on and you have the circuit below:







simulate this circuit

Adding up the currents as shown, and rearranging, you can find an expression for \$V_{out}\$ in terms of \$V_{in}\$.

Putting it all together

So in essence, you have three different circuits when you sweep \$V_{in}\$: 

$$V_{in} &lt; -(V_{D2} + E), \quad \text{D1 off, D2 on}, \quad V_{out} = -(V_{D2} + E) $$
$$-\frac{R_2 + R_3}{R_2}(V_{D2}+E) &lt; V_{in} &lt; \frac{R_2 + R_3}{R_3} V_{D1}, \quad \text{D1 off, D2 off} \quad V_{out} = \frac{R_2}{R_2 + R_3} V_{in} $$
$$ V_{in}&gt;\frac{R_2 + R_3}{R_3} V_{D1}, \quad \text{D1 on, D2 off} \quad V_{out} = (R_1 || R_2 || R_3)(\frac{V_{in}}{R_1 || R_3}-\frac{V_{D1}}{R_1})$$

Now you will get different results if you assume D1 and D2 are different diodes. Also, when you are doing these kinds of problems, as a sanity check you should make sure your \$V_{out}\$ graph is continuous by checking the boundary conditions.