Step 1:

Although [latex]\infty[/latex] is not really a number, we can still try plugging it in and seeing how the function behaves. So, when substituting, we have:

[latex]\large{\lim_{x \to -\infty}\frac{e^x+2x}{e^x-3x} = \frac{e^{-\infty}+2(-\infty)}{e^{-\infty}-3(-\infty)} = \frac{\infty}{\infty}}[/latex]

Thus, we are left with an indeterminate form. This means that we must rewrite the function, before plugging in the limit.

Step 2:

When rewriting polynomial and rational functions, one useful technique we can use is to factor out the highest degree from both the numerator and denominator. In this case, the highest degree is [latex]x[/latex]. So, we have:

[latex]\large{\lim_{x \to -\infty}\frac{e^x+2x}{e^x-3x} = \lim_{x \to -\infty}\frac{x(\frac{e^x}{x}+2)}{x(\frac{e^x}{x}-3)}}[/latex]

[latex]\large{ = \lim_{x \to -\infty}\frac{\frac{e^x}{x}+2}{\frac{e^x}{x}-3}}[/latex]

Step 3:

Now, using the limit laws, we can evaluate the limit for each term as shown:

[latex]\large{ = \lim_{x \to -\infty}\frac{\frac{e^x}{x}+2}{\frac{e^x}{x}-3} = [latex]\large{ \frac{\lim_{x\to -\infty}(\frac{e^x}{x})+\lim_{x\to -\infty}(2)}{\lim_{x\to -\infty}(\frac{e^x}{x})-\lim_{x\to -\infty}(3)}}}[/latex]

Hence, we have:

[latex]\large{ = \frac{0+2}{0-3} = -\frac{2}{3}}[/latex]

Thus, the final answer is:

[latex]\large{\lim_{x \to -\infty}\frac{e^x+2x}{e^x-3x} = -\frac{2}{3}}[/latex]