1. ## L'Hospital's Rule True/False

Hi, can anyone tell me which of these are wrong?

1. The indeterminate form of type "0/0" can approach any real number as a limit -- False

2. There is a natural number n such that the power function f(x)=x^n is growing faster than the exponential function g(x)=e^x -- True

3. if lim x->infinity f(x) = +infinity, then for any k (all real numbers), lim x->+infinity k/f(x) = 0 -- False

4. The function f(x) = lnx grows more slowly than any positive power of x -- True

Thanks!

2. Originally Posted by coldfire
Hi, can anyone tell me which of these are wrong?

1. The indeterminate form of type "0/0" can approach any real number as a limit -- False
ok

3. if lim x->infinity f(x) = +infinity, then for any k (all real numbers), lim x->+infinity k/f(x) = 0 -- False
not ok
why do you say it's false ?

2. There is a natural number n such that the power function f(x)=x^n is growing faster than the exponential function g(x)=e^x -- True

4. The function f(x) = lnx grows more slowly than any positive power of x -- True
the exponential grows faster than any power of x, which grows faster than the logarithm.

3. disagree with #1 if the limit exists.

4. Originally Posted by coldfire
Hi, can anyone tell me which of these are wrong?
1. The indeterminate form of type "0/0" can approach any real number as a limit -- False
True, think about $\lim_{x\to{0}}\frac{1-\cos(x)}{x^2}$

This yields $\frac{1}{2}$

So if $a\in\mathbb{R}$ then $\lim_{x\to{0}}\frac{a-a\cos(x)}{x^2}=\frac{a}{2}$

Supposing the limit exists of course.
2. There is a natural number n such that the power function f(x)=x^n is growing faster than the exponential function g(x)=e^x -- True
No! Consider that the function $x^n$ can be differentiated and n+1 number of times to give $n!$ and if you differentiate $e^x$ an n+1 number of times you get $e^x$ and it is pretty obvious that

$\forall{n}\in\mathbb{N}~\exists{N}\backepsilon\for all{x}\geqslant{N}~n!\leqslant{e^x}$
3. if lim x->infinity f(x) = +infinity, then for any k (all real numbers), lim x->+infinity k/f(x) = 0 -- False
True! \begin{aligned}\lim_{x\to\infty}\frac{k}{f(x)}&=k\ lim_{x\to\infty}\frac{1}{f(x)}\\
&=k\cdot{0}\left({\color{red}\text{Why?}}\right )\\
&=0\end{aligned}

4. The function f(x) = lnx grows more slowly than any positive power of x -- True
Yes! The reason being that $\left(\ln(x)\right)'=\frac{1}{x}\leqslant{\varphi{ x}^{\varphi-1}}~\varphi>0$

5. Originally Posted by skeeter
disagree with #1 if the limit exists.
Originally Posted by Mathstud28
True
What's disturbing me for the first one is that it is written "can reach any real value"

so what I understand is that since it is an indeterminate form 0/0, we can attribute any value we want to the limit. Which is not true.
Can someone explain it more clearly ?

6. Originally Posted by Moo
What's disturbing me for the first one is that it is written "can reach any real value"

so what I understand is that since it is an indeterminate form 0/0, we can attribute any value we want to the limit. Which is not true.
Can someone explain it more clearly ?
Does this help? Suppose $\lim_{x\to{c}}\frac{f(x)}{f(c)}=L$ where $f(c)=g(c)=0$

Then the limit $\lim_{x\to{c}}\frac{\gamma{f(x)}}{g(x)}=\gamma{L}~ \gamma\in\mathbb{R}$

Note this would still be a "0/0" form since $\gamma{f(c)}=\gamma{0}=0~\gamma\in\mathbb{R}$

Now let $\varphi\in\mathbb{R}$, let this be the "any value" desired, then all thats left to do is solve $\gamma{L}=\varphi$