Calculus question...calculus experts please...?

Question

how do i solve lim h->0 (e^h-1)/3h , the answer is 1/3 but how do i get it? also, it is e as in 2.718...thanks.

Yeah..I know about the hospital rule but I'm not supposed to use it yet! Our class hasn't learned it yet lol.

Answer 1

Yes, it's e as in 2.718.

If you plug in h=0, you get 0/0. So apply l'Hospital's Rule and differentiate the numerator and denominator.

you get lim h->0 e^h / 3.

Plug in h=0 and you get 1/3.

Answer 2

This lim is inderterminate (0/0) but you have precluded the use of L'Hopital's Rule. So.....

e^x = 1 + x + x^2/2 + ........
As x goes to 0 then e^x goes to 1 + x

Therefore your limit becomes

lim (1 + h - 1)/3h = 1/3 lim 1 = 1/3

with all the above limits as h tends to 0.

Go in peace....

Answer 3

L'hopital's rule is very intuitive. Even if you've never heard of it before, you can come up with the idea logically (or at least, I did it in G.12). Say you have two functions f(x), g(x) that meet at a point (a,b). The limiting ratio of their heights from b is the same as the ratio of their slopes at (a,b). Algebraically,

Lim{x->a} [f(x)-b] / [g(x)-b] = f'(a) / g'(a)

This is a direct result of the "smooth functions look like straight lines up close" idea. This can be applied recursively.

Let's apply this to your problem.

Lim {x->0} (e^x-1)/(3x)
=(1/3) Lim {x->0} (e^x-1)/x

Notice the two functions e^x-1 and x meet at (0,0). Notice, from definition of e^x, e^x-1 has slope 1 at 0. Since both functions have slope 1 at the meeting point, the ratio is 1. (1/3)*(1) = 1/3

Answer 4

Direct substitution yields (1-1)/(3*0) = 0/0, an indeterminate form. Therefore, apply L'hopital's rule (where f(x)→0 and g(x)→0 as x→c, [x→c]lim f(x)/g(x) = [x→c]lim f'(x)/g'(x) ):

[h→0]lim (e^h-1)/(3h) = [h→0]lim e^h/3 = 1/3

Answer 5

Without using LHopital:
The derivative of e^h at zero is
Lim (e^h - 1) / h = (e^h)'(0) = (e^h)(0) = 1
So
Lim (e^h - 1) / 3h = 1/3 Lim (e^h - 1) / h = 1 / 3

Answer 6

There are two ways to solve this

because you know calculus you recognize the left hand side as

1/3(lim h->0 (e^h -1)/h)
= 1/3 lim h->0 (e^(x+h) - e(x))/h when x =0 e^ x= 1
= 1/3 d/dx(e^x) at 0

= 1/3 e^x at x =0 = 1/3.

secondly you can use series expansion and do it as others have mentioned .

Answer 7

lim h->0 (e^h-1)/3h if you substitute h=0 you get
(1-1)/0 which is undefined, so use L-hospitals rule:
let f(h)=e^h-1 & g(h)=3h
lim h-->0 f(h)/g(h)=lim h---> f'(h)/g'(h)
f'(h)=e^h; g'(h)=3
f'(0)/g'(0)=e^0/3=1/3

Answer 8

I forget this stuff, but I have a feeling what they say isn't helpful because at this point you definitely would not have covered l'hopitals rule because I remember not even getting to l'hopitals rule in my calc class...they probably want you to do it some other way

Answer 9

Have you had the definition of the exponential as an infinite sum?

e^x = 1 + x + x^2/2! + x^3/3! + ... for |x| < 1.

If you use this it is easy: e^x - 1 = x + x^2/2! + x^3/3! + ...
(e^x - 1)/(3x) = 1/3 + x/(3*2!) + x^2/(3*3!) + ...
= 1/3

Answer 10

lim as x-> 0 of f(x) = constant, means f(anything)=constant. answer: a^3