L'Hopital's Rule

[jtw2e2]

lim = (5[sup:2h2h06bz]t[/sup:2h2h06bz]-3[sup:2h2h06bz]t[/sup:2h2h06bz])/t
t->0

Using L'Hopital's Rule, how can one differentiate this? Won't you end up with (t*5[sup:2h2h06bz]t-1[/sup:2h2h06bz] - t*3[sup:2h2h06bz]t-1[/sup:2h2h06bz])/1 ???

 

[daon]

You used the power rule. You are not raising a function of t to a real power, which is when the power rule applies. I found that this is a common mistake among students in Calculus.

The derivative of a^t w.r.t. to t is a^t * ln(a). You would use the power rule if you were given t^a, not a^t.

 

[BigGlenntheHeavy]

$\displaystyle \lim_{t\to0}\frac{5^t-3^t}{t} \ gives \ one \ the \ indeterminate \ form \ \frac{0}{0}, \ hence \ the \ Marqui \ to \ the \ rescue.$

$\displaystyle D_t \ [5^t-3^t] \ = \ 5^tln5-3^tln3 \ and \ D_t \ [t] \ = \ 1$

$\displaystyle Hence \ we \ have \ \lim_{t\to0} \ [5^tln5-3^tln3] \ = \ ln5-ln3 \ = \ ln(5/3).$

$\displaystyle Look, \ D_t[t^2] \ = \ 2t, \ exponent \ a \ constant, \ but \ D_t[2^t] \ = \ 2^tln|t|, \ exponent \ a \ variable.$

$\displaystyle D_t[t^2] \ the \ hard \ way. \ Let \ y \ = \ t^2, \ then \ ln|y| \ = \ 2ln|t|, \ y \ = \ e^{2ln|t|}$

$\displaystyle y' \ = \ \frac{2e^{2ln|t|}}{t} \ = \ \frac{2t^2}{t} \ = \ 2t, \ Note: \ t^2 \ = \ e^{2ln|t|}$

$\displaystyle Now, \ D_t[2^t] \ Again \ let \ y \ = \ 2^t, \ then \ lny \ = \ tln(2) \ \implies \ y \ = \ e^{tln(2)}$

$\displaystyle Ergo, \ y' \ = \ ln(2)e^{tln2} \ = \ 2^tln(2), \ Again \ note: \ e^{tln(2)} \ = \ 2^t.$

$\displaystyle Hope \ this \ helped \ as \ it \ is \ a \ little \ confusing \ at \ first.$

 

[jtw2e2]

You used the power rule. You are not raising a function of t to a real power, which is when the power rule applies. I found that this is a common mistake among students in Calculus.

The derivative of a^t w.r.t. to t is a^t * ln(a). You would use the power rule if you were given t^a, not a^t.

Thank you so much. That is assuming "a" is a constant, right?

 

[jtw2e2]

Thanks Glenn, but how does one go from here:

$\displaystyle [5^tln5-3^tln3]$

to here:

\(\displaystyle \ = \ ln5-ln3 \\)

If 5[sup:c4geqw6p]t[/sup:c4geqw6p] is a constant multiplier, doesn't it have to stay? I understand if it were a constant added to ln5 its derivative would simply be 0 but since it is multiplied, we can't just drop it, can we?

 

[BigGlenntheHeavy]

$\displaystyle Anything \ raised \ to \ the \ zero \ power \ is \ 1, \ except \ 0^0.$

$\displaystyle Note: \ Easy \ proof, \ \frac{x^1}{x^1} \ = \ 1 \ = \ x^{1-1} \ = \ x^0, \ hence \ x^0 \ must \ = \ 1.$

$\displaystyle \lim_{t\to0} \ (5^t) \ = \ 1$

 

[jtw2e2]

$\displaystyle Anything \ raised \ to \ the \ zero \ power \ is \ 1, \ except \ 0^0.$

$\displaystyle Note: \ Easy \ proof, \ \frac{x^1}{x^1} \ = \ 1 \ = \ x^{1-1} \ = \ x^0, \ hence \ x^0 \ must \ = \ 1.$

$\displaystyle \lim_{t\to0} \ (5^t) \ = \ 1$

Thank you.

 

[daon]

It is okay to define 0^0 to be 1; many calculators do this. IMO, It makes about the same sense as defining anything to the 0th power to be 1. In group theory, g^0 is often called the "empty product" and is defined to be the multiplicative identity. I'm open to any objections, as I don't recall any reason why not to define it that way.

 

[BigGlenntheHeavy]

$\displaystyle daon ,\ I'll \ pass \ as \ I \ have \ been \ drawn \ into \ this \ debate \ to \ many \ times \ to \ no \ avail.$

$\displaystyle However, \ I'm \ curious, \ what \ does \ IMO \ mean?$

 

[daon]

Sorry, too much forum posting recently. It's an abbreviation for "in my opinion."