# Math Help - L'Hopital's Rule...!

1. ## L'Hopital's Rule...!

Determine if L'Hopital's rule applies in each case. Select all that apply.

lim_y-->0 3^y/y^2

lim_x-->pi (sin(3x))/(x-pi)

lim_t-->0 (t^2+3t)/(cosh(t)-1)

lim_z-->0 (e^(2z)-1)/e^z

lim_theta-->0 (arctan(theta))/(3(theta))

lim_x-->infinity (e^(-x))/(1+ln(x))

lim_x-->0+ (cot(x))/(ln(x))

lim_t-->infinity (ln(t))^2/t

we haven't exactly gone over L'Hopital's rule yet, but im tryin to do it early, any help on how you can tell whether the rule will work or not, thanks...

mathete

2. l'hospital's rule:

if:
$\lim f(x) =0\ and\ \lim g(x) =0$

or if:
$\left| \lim f(x) \right| =\infty\ and\ \left| \lim g(x) \right| =\infty$

then:
$\lim \frac{f(x)}{g(x)}=\lim \frac{f'(x)}{g'(x)}$

For your questions, just make sure that the limits of the quotients are of the forms $\frac{0}{0}$ or $\frac{\infty}{\infty}$ as described above.

3. ok, so does that just mean that the if the limit of some function is 0 or infinity then l'hopital's rule applies?

4. Not quite, it requires that the limit of the quotient of functions (you'll notice that all your examples are in or can be written in the form $\frac{f(x)}{g(x)}$) must be one of the indeterminate forms: $\frac{0}{0}$ or $\frac{\infty}{\infty}$ (0r $-\frac{\infty}{\infty}$). If the limit of the quotient were $\frac{0}{\infty}$, or $\infty$ or 2 or anything else, then L'hospital does not apply.

5. OK, do you think you could show me an example of one of the ones that i put up there...that would help, thanks

6. maybe this will clear things up...

$\lim_{x\rightarrow 0} \left(\frac{t^2+3t}{\cosh(t)-1}\right)=\frac{\lim_{x\rightarrow 0} (t^2+3t)}{\lim_{x\rightarrow 0}(\cosh(t)-1)}=\frac{0}{0}\Rightarrow$ L'Hospital's law applies

$\lim_{x\rightarrow 0^{+}} \left(\frac{\cot(x)}{\ln(x)}\right)=\frac{\lim_{x\ rightarrow 0^{+}}(\cot x)}{\lim_{x\rightarrow 0^{+}}(\ln x)}=\frac{\infty}{-\infty}\Rightarrow$ L'Hospital's law applies

$\lim_{x\rightarrow \infty} \left(\frac{e^{-x}}{1+\ln x}\right)=\frac{\lim_{x\rightarrow \infty} e^{-x}}{\lim_{x\rightarrow \infty} 1+\ln x}=
\frac{0}{\infty}\Rightarrow$
L'Hospital's law does not apply

7. Hello, mathlete!

ok, so does that just mean that the if the limit of some function
is 0 or infinity then l'hopital's rule applies?
I wouldn't put it quite like that . . .

If the limit (as written) is an indeterminate form, then L'Hopital applies.

There are several indeterminate forms.

The two basic forms are: . $\frac{0}{0}\text{ and }\frac{\infty}{\infty}$ . . . and L'Hopital can be applied.

There are other indeterminate forms: . $\infty - \infty,\;0^0,\;0^{\infty},\;\infty^0,\;1^{\infty}, \;\hdots$
Often these can be transformed into one of the two basic forms,
. . then L'Hopital can be used.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

$1)\;\;\lim_{y\to0}\frac{3^y}{y^2} \;=\;\frac{3^0}{0^2} \;=\;\frac{1}{0}$ . . . not indeterminate

$2)\;\;\lim_{x\to\pi}\frac{\sin(3x)}{x-\pi} \;=\;\frac{\sin(3\pi)}{\pi-\pi} \;=\;\frac{0}{0}$ . . . yes, indeterminate!

$3)\;\;\lim_{t\to0} \frac{t^2+3t}{\cosh t - 1} \;=\;\frac{0^2 + 3(0)}{\cosh(0) - 1} \;=\;\frac{0+0}{1-1} \;=\;\frac{0}{0}$ . . . yes, indeterminate!

$4)\;\;\lim_{z\to0}\frac{e^{2x}-1}{e^x} \;=\;\frac{e^0-1}{e^0} \;=\;\frac{1-1}{1} \;=\;\frac{0}{1} \;=\;0$ . . . not indeterminate

Get the idea?