When working with a limit of e^x such as:

The limit as x approaches infinity of

Can I simply state that it tends towards 1/((e^inf)/inf) and therefore to 0, or is the limit not yet defined since it tends towards inf/inf below 1? Thanks.

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- October 12th 2013, 07:03 AMPazeLimit of e^x
When working with a limit of e^x such as:

The limit as x approaches infinity of

Can I simply state that it tends towards 1/((e^inf)/inf) and therefore to 0, or is the limit not yet defined since it tends towards inf/inf below 1? Thanks. - October 12th 2013, 07:50 AMemakarovRe: Limit of e^x
Why do you think that 1/((e^inf)/inf) is 0? In general, you can't compare infinities.

- October 12th 2013, 11:39 AMPazeRe: Limit of e^x
- October 12th 2013, 11:49 AMemakarovRe: Limit of e^x
Yes, you could use L'Hopital's rule. To find limits, you need some database of standard facts, e.g., as provided and . The fact that the where is one of those standard facts, but one has to prove it for the first time.

- October 12th 2013, 03:05 PMHallsofIvyRe: Limit of e^x
You could also use the MacLaurin series for , so that .

Now you have .

Divide both numerator and denominator by to get which goes to infinity as x goes to infinity. Notice that, by this argument, . - October 12th 2013, 03:47 PMPazeRe: Limit of e^x
- October 12th 2013, 03:52 PMPlatoRe: Limit of e^x
- October 12th 2013, 06:08 PMPazeRe: Limit of e^x
- May 5th 2015, 08:26 PMlimiTSRe: Limit of e^x
I think you cannot use the Maclaurin series for $e^x$ which is about $x=0$ to analyze $e^x$ when $x\to\infty$. The expansion point (zero) is too far away from the limit point (infinity).

To solve the problem, you have to use L'Hospital's or just know that exponentials grow faster than polynomials.

This statement above is false: $\displaystyle\lim_{x\to\infty}\frac{x}{1-e^x}=-1$. The Maclaurin series approach gives an erroneous answer, because it cannot be used here.

Rather, the limit equals 0, as pointed out in other posts. - May 5th 2015, 09:07 PMArchieRe: Limit of e^x
$$\begin{aligned}

0 &\lt {1 \over t} \lt t^{c-1} \qquad (t \gt 1,\, c \gt 0) \\

\int_1^x 0 \,\mathrm d t &\lt \int_1^x {1 \over t} \,\mathrm d t \lt \int_1^x t^{c-1} \,\mathrm d t \qquad (x \gt 1) \\

0 &\lt \log x \lt {x^c - 1 \over c} \lt {x^c \over c} \\

0 &\lt \log^a x \lt {x^{ac} \over c^a} \qquad (a \gt 0) \\

0 &\lt {\log^a x \over x^b} \lt {x^{ac-b} \over c^a} \qquad (b \gt 0) \\

\end{aligned}$$

Since this is true for all $c \gt 0$ we may pick any $c = {b \over 2a}$ so that

$$

0 \lt {\log^a x \over x^b} \lt {x^{-{b \over 2}} \over c^a} \\

$$

And now, taking the limit as $x \to \infty$ gives

$$

0 \le \lim_{x \to \infty} {\log^a x \over x^b} \le \lim_{x \to \infty} {x^{-{b \over 2}} \over c^a} = 0 \\ \implies \lim_{x \to \infty} {\log^a x \over x^b} = 0

$$

Then, writing $x = \mathrm e^y$ we have

$$\begin{aligned}

\lim_{x \to \infty} {\log^a x \over x^b} &= 0 \\

\lim_{y \to \infty} {\log^a \mathrm e^y \over \mathrm e^{by}} &= 0 \\

\lim_{y \to \infty} {y^a \over \mathrm e^{by}} &= 0 \\

\end{aligned}$$

Now, addressing your question

$$\begin{aligned}

\lim_{x \to \infty} {x^2 \over 1 - \mathrm e^x} &= \lim_{x \to \infty} {x^2 \mathrm e^{-x} \over \mathrm e^{-x} - 1} \\

&= \lim_{x \to \infty} {{x^2 \over \mathrm e^{x}} \over \mathrm e^{-x} - 1} \\

&= {0 \over 0 - 1} = 0

\end{aligned}$$ - May 5th 2015, 10:07 PMvotanRe: Limit of e^x
Is it

*x*--> + infinity or - infinity. You could alternatively reason this way:

as*x*--> + infinity (1 - e^x) --> (- e^x)

At the limit the ratio varies as x^2/(-e^x) and the limit is 0 from the left

as*x*--> - infinity (1 - e^x) --> 1

At the limit the ratio varies as x^2 --> + infinity