lim of exponential funct

[calcstruggles2013]

This is the problem:

[SIZE=-1][/SIZE] lim [SIZE=-1]h→ 0[/SIZE]

43.6h −1 h

1. I tried using L'Hopitals Rule, because if you sub in zero into this prob you are stuck with "0/0".
basically I got 4.36^h * ln43.6 and when I subbed in zero, I got zero, but lon capa (math hw web site) said that was incorrect.
I used this method bc in my book it says that (a^x)'= a^x * lnx but there is a chance I am not applying it correctly.
I also tried "1" as the answer bc my teacher mentioned that all exponential functions pass through the point (0,1) but lon capa did not accept that either.
I also tried this problem using a similar thought process.

lim [SIZE=-1]h→ 0[/SIZE]

1.5−1+h −1.5−1 h

and I got -->
ln(1.5)*(-1.5)^-2+ln(1.5)*(1.5)^-2

 

[srmichael]

This is the problem:

lim [SIZE=-1]h→ 0[/SIZE]

43.6h −1 h

1. I tried using L'Hopitals Rule, because if you sub in zero into this prob you are stuck with "0/0".
basically I got 4.36^h * ln43.6 and when I subbed in zero, I got zero, but lon capa (math hw web site) said that was incorrect.
I used this method bc in my book it says that (a^x)'= a^x * lnx but there is a chance I am not applying it correctly.
I also tried "1" as the answer bc my teacher mentioned that all exponential functions pass through the point (0,1) but lon capa did not accept that either.
I also tried this problem using a similar thought process.

lim [SIZE=-1]h→ 0[/SIZE]

1.5−1+h −1.5−1 h

and I got -->
ln(1.5)*(-1.5)^-2+ln(1.5)*(1.5)^-2

You are correct in your derivative. It is equal to (43.6^h)ln(43.6), but when you plug in 0 for h it does not equal 0. It equals ln(43.6) since 43.6^0 is 1. So the answer to your first limit is ln(43.6).

 

[lookagain]

lim [SIZE=-1]h→ 0[/SIZE]

43.6h −1 h

...my teacher mentioned that all exponential functions pass through the point (0,1)
I also tried this problem using a similar thought process.

lim [SIZE=-1]h→ 0[/SIZE]

1.5−1+h −1.5−1 h

and I got -->
ln(1.5)*(-1.5)^-2+ln(1.5)*(1.5)^-2

Not all exponential functions pass through (0, 1).


2nd problem:


In some order, the derivative of the numerator is equal to

\(\displaystyle [1.5^{-1 + h}]\bigg[\dfrac{d}{dh}(-1 + h)\bigg][ln(1.5)] \ - \ 0 \ = \)

\(\displaystyle [ln(1.5)][1.5^{-1 + h}]\)


What happens with that expression when you take the limit as \(\displaystyle h \to 0 \ ?\)

 

[HallsofIvy]

You might also recognize the original limit, \(\displaystyle \lim_{h\to 0}\frac{43.6^h- 1}{h}\) as being the definition of the derivative: \(\displaystyle \lim_{h\to 0}\frac{f(a+ h)- f(a)}{h}\) with \(\displaystyle f(x)= 43. 6^h\), and a= 0. Naturally, when you use "L'hopital's rule" the derivative of the denominator is 1 so you are just evaluating the derivative of \(\displaystyle 43.6^h\) at x= 0 which is \(\displaystyle (43.6^0)ln(43.6)= ln(43.6)\) since, as srmichael said, \(\displaystyle a^0= 1\) for any positive a.