I am working with vector functions and one of the problems has $e$ in it. I always thought that $$\lim_{x\to\infty}e^{ax} = \infty$$ but the book says otherwise and when I went all the way back to chapter 2 there was a question:
$$\lim_{x\to\infty} \frac{1-e^{x}}{1+2e^x} = -\frac{1}{2}$$ How is that case? If e goes to infinity? even applying l'hospitals rule the $e^{x}$ doesnt go away what am I missing here? So you can see the two in perspective here is the problem that I am working with involving vectors:
$$\lim_{t\to\infty} \frac{1-e^{-2t}}{t}$$
What am I forgetting about limits?
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$\begingroup$For the first limit it'll have to depend on what the value of "a" is. If a is nonpositive, as you can see, the limit will be 0. And for the second limit, after applying L'hospitals' rule, I believe you will only have -e^x/2e^x that simplifies to -1/2, so e^x should go away.
$\endgroup$ 2 $\begingroup$First off, note that $$ \lim_{x\rightarrow\infty}e^{ax}=\begin{cases} \infty & \text{if }a>0;\\ 1 & \text{if }a=0;\\ 0 & \text{if }a<0. \end{cases} $$
Now, this has not much to do with the limit you mention. Also, as K.Gibson points out, $e$ is not the variable going to infinity ($e$ is just a constant!).
L'Hopital's rule gives
$$\lim_{x\rightarrow\infty}\frac{1-e^{x}}{1+2e^{x}}=\lim_{x\rightarrow\infty}\frac{-e^{x}}{2e^{x}}=-\frac{1}{2}$$
$\endgroup$ $\begingroup$If $a < 0$ $\lim_{x \rightarrow} e^{ax} = 0$.
If $a = 0$ $\lim_{x \rightarrow} e^{ax} = 1$.
If $a > 0$ $\lim_{x \rightarrow} e^{ax} = \infty$.
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L'hopital: $\lim \frac{1 - e^x}{1 + 2e^x} = \lim \frac{-e^x}{2e^x} = - \frac 12$.
Not using L'hopital: $\lim \frac{1 - e^x}{1 + 2e^x}=\lim \frac{1/e^x - 1}{1/e^x + 2}= -\frac 12$
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