# Limit product

1. Jul 2, 2013

### scientifico

Hello, why $lim_{x\to (0)^{+}}e^{1/x}3x^2 = +\infty$

if

$lim_{x\to (0)^{+}}e^{1/x} = \infty$ and $\lim_{x\to (0)^{+}}{3x^{2}} = 0$

shouldn't it be $+\infty * 0$ ? I can't get it :(

Thanks

2. Jul 2, 2013

### WannabeNewton

$\infty*0$ is not $0$; this is what we call an indeterminate form. Intuitively you can figure out the limit by looking at which of the two elementary functions in the limit grows or decreases faster. $e^{1/x}$ goes to $\infty$ much faster than $3x^{2}$ goes to zero.

3. Jul 2, 2013

### jbunniii

Consider the Taylor series for the exponential function:
$$e^{x} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$$
This is valid for all $x$. Therefore, for $x \neq 0$, we have
$$e^{1/x} = 1 + \frac{1}{x} + \frac{1}{2!x^2} + \frac{1}{3!x^3} + \cdots$$
and so
$$3x^2 e^{1/x} = 3x^2 + 3x + \frac{3}{2} + \frac{1}{2x} + \cdots$$
In this last expression, all of the terms to the right of 3/2 will have positive powers of $x$ in the denominator, so they will blow up to $+\infty$ as $x \rightarrow 0^+$. Meanwhile, the first two terms go to zero, and the third term is just 3/2. Therefore the overall limit is $0 + 0 + 3/2 +\infty = +\infty$.

Last edited: Jul 2, 2013
4. Jul 2, 2013

### scientifico

I still can't understand Taylor series, I just would like to know why in this case the limit product rule doesn't work (ok for the infinity grade but shouldn't it work with as a product too ? )

5. Jul 2, 2013

### Stephen Tashi

What is the statement of the "limit product" rule in you study materials? Doesn't it say that both limits must exist (and thus be finite)? Perhaps the relevant question is why you would expect it work in a situation where it hasn't been proven to apply.

6. Jul 3, 2013

### scientifico

I mean $\lim_{x\to c}{f(x)\cdot g(x)} = \lim_{x\to c}{f(x)}\cdot\lim_{x\to c}{g(x)}$

Why they must be finite and anyway what allow me to ignore the other limit in case it isn't ?

Last edited: Jul 3, 2013
7. Jul 3, 2013

### WannabeNewton

That only holds if the limits actually exist.

8. Jul 3, 2013

### jbunniii

A "limit product" of the form $\infty \cdot 0$ is ill-defined. The limit may be $0$, $\infty$, or any finite number, or it may not exist at all.

Here are four examples where $\lim_{x \rightarrow \infty} f(x) = \infty$ and $\lim_{x \rightarrow \infty} g(x) = 0$, but $\lim_{x \rightarrow \infty} f(x)g(x)$ yields different answers:

1. Let $f(x) = x^2$ and $g(x) = 1/x$. Then $\lim_{x \rightarrow \infty} f(x) g(x) = \lim_{x \rightarrow \infty} x = \infty$.

2. Let $f(x) = x^2$ and $g(x) = 1/x^3$. Then $\lim_{x \rightarrow \infty} f(x) g(x) = \lim_{x \rightarrow \infty} 1/x = 0$.

3. Let $c > 0$ and $f(x) = cx$ and $g(x) = 1/x$. Then $\lim_{x \rightarrow \infty} f(x) g(x) = \lim_{x \rightarrow \infty} cx/x = c$.

4. Let $f(x) = x$ and $g(x) = \sin(x)/x$. Then $\lim_{x \rightarrow \infty} f(x) g(x) = \lim_{x \rightarrow \infty} \sin(x)$ does not exist.

9. Jul 3, 2013

### micromass

Staff Emeritus
Awesome post...

10. Jul 3, 2013

### scientifico

In that cases you simplified for x, how can I rewrite $e^{1/x}3x^2$ to simplify it ?

11. Jul 3, 2013

### jbunniii

You can't simplify it in the way I did above. You need to somehow use the fact that the exponential function grows faster than any polynomial function, in the sense that
$$\lim_{x \rightarrow \infty} \frac{e^x}{x^n} = \infty$$
for any positive integer $n$. The Taylor series argument I gave above implicitly includes this fact by (loosely speaking) representing the exponential function as a polynomial with "infinite degree", but there are other ways to prove it as well. Any rigorous argument will depend on how you have defined $e^x$. The series representation is one way to define it. What is the definition you are working with?