Solve the following limit---(e^x-1)/x as x approaches zero, without L'hopital's rule.?

Question

With steps...

Mr. im not in high school, and this is no HW. i was just curious, how do they solve this, as i wasnt able to find the proof anywhere.

ok if we use L'hoptital's rule then do u know that we find the derivative of e^x by this limit only.

Ya thats true we can use taylor series for e^x. but we derive the series by using the derivatives of e^x. and that we calculate by this limit.so u dont solve the fundamental by the derived thing.

Dont u people understand that to use L'hopital's rule we need the derivative of e^x which is calculated by this limit. how can u solve this with what u have derived from it.

Answer 1

This is the *definition* of the derivative of e^x at x=0. Since you know the derivative of e^x is e^x, let x=0 to get the limit is 1.

Now, you might ask why we know that the derivative of e^x is again e^x. That is again almost be definition. We know, through the limit definition of the derivative, that the derivative of a^x is a constant times a^x for any fixed number 'a'. For a=2, we can check directly that the constant is less than 1 and for a=3 that the constant is more than 1. We *define* e to be that value of 'a' for which the constant is equal to 1. That gives the derivative as we want it.

Answer 2

E X 1 X

Answer 3

The answer before mine is the proper and elegant way to solve the limit without L'Hopital's rule.

Remember that e^x is a continuous, real function which is defined for all real numbers x - the limit of e^x as x approaches - infinity is zero.

The derivative of e^x is itself - so its derivative is real and continuous. All derivatives of e^x are real and continuous as well.

Answer 4

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If a is a positive constant, d/dx(a^x)=a^x * ln(a). 1. Using definition of the derivative, f'(c)=\lim_{x \to c} (f(x)-f(c))/(x-c) with f(x)=a^x and c=0 gives us ln(a)=\lim_{x \to 0} (a^x-1)/x 2. Using L'Hospital's rule, \lim_{x \to 0} (a^x-1)/x =\lim_{x \to 0} d/dx( (a^x-1) )/ d/dx( x ) =\lim_{x \to 0} a^x * ln(a)/1 =ln(a). PS To see that the derivative of a^x is a^x ln (a). y=a^x Take the natural logarithm of both sides: ln(y)=x ln(a) Implicitly differentiate with respect to x: 1/y y'=ln(a) [remember, ln(a) is a constant.] Solve for y' y'=y ln(a) But, y=a^x so y'=a^x ln(a)

Answer 5

Yes, Mathematician is right. It's definitional. This is why it's called the "natural" base. It's the base for which the unnits of time in the usual differential equation

df/dt=kf

are taken such that k=1. The number you get by (for example) solving this iteratively and taking the limit is
(something)^t
where t is measured in the "natural units." The (something) is what we call "e".

Answer 6

I've put a beautifully formatted solution at the link below. Take a look.
http://www.tomsmath.com/limit-of-e-to-the-x-minus-1-over-x-as-x-goes-to-zero.html

Answer 7

As everyone has stated, this is a "typical" L'Hospital's rule problem or observed that

d/dx(e^x) at x=0 is

\lim_{x \to 0} (f(x)-(0))/(x-0)=(e^x-1)/x.

Since the derivative of e^x is e^x, its value at x=0 is 1 so the limit is 1.

We can also do it other ways without L'Hospital's rule.

Using series we know that e^x=\sum_{k=0}^{\infty} x^k/k! so

e^x-1=\sum_{k=1}^{\infty} x^k/k!

Thus,

(e^x-1)/x=(\sum_{k=1}^{\infty} x^k/k!)/x
=(\sum_{k=1}^{\infty} x^(k-1)/k!)
=1+x/2!+x^3/3!+...

Thus,

\lim_{x \to 0} (e^x-1)/x
= \lim_{x \to 0} (1+x/2!+x^3/3!+...)=1.

You should try computing \lim_{x \to 0} \sin(x) / x in the same ways! (That is using series, and then using the fact that \lim_{x \to 0} \sin(x) / x is the derivative of sin(x) at x=0: d/dx (sin(x))=cos(x) and cos(0)=1.) :)

Answer 8

Evaluate the function at x = 0.1 and x = - 0.1. Then evaluate the function at x = 0.01 and x = - 0.01. Keep doing this, making x smaller by a factor of 10^-1 each time, and you will see that the function approaches 1 from both sides.

So the limit is 1

Answer 9

What a classic problem

Just remember that e^x = sum(x^k/k!) for k = 0 to ∞.

Now make the substitution x-1 for x in the expansion, work out the first few terms, and you'll see the pattern.

Then do the arithmetic


Doug

Answer 10

u would hav to do it with l'hopital's rule. and it would be:

limit as x approaches 0 ---> derivative of (e^x-1)/derivative of x

limit as x approaches 0 ---> e^x/1

answer is 1