Solve the limit --- (a^x-1)/x as x approaches zero.?

Question

in the (e^x-1)/x limit u people talked about definition, ok then solve this limit!!! now again u will say, use l'hopital's rule, then again i will say how can u use it here if u find the derivative of a^x by this limit. now dont say its the definition of a^x. plz

we really have bright people around here. TUBE LIGHT.

ok then mr. plz calculate the derivative of a^x with respect to x, without this limit. if that happens then surely i will use l'hopital's rule to solve this limit.

Answer 1

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 2

lim x->0 (a^x-1)/x can be found using L'Hospital's Rule as you state. Let f(x) = a^x - 1 and g(x) = x, then you seek lim x->0 f(x)/g(x). Note: the derivative of a^x is ln a * a^x.

L'Hospital's Rule states that as f(x) -> 0 and g(x) -> 0 as x -> 0 then lim x->0 f(x)/g(x) = lim x->0 f'(x)/g'(x).

Hence lim x->0 (a^x-1)/x = lim x->0 ln a * a^x = ln a.

It has been assumed here that a is a positive real number.

Answer 3

If you want to differentiate it, then here is my calculations:
Given:
(a^x -1)/x

Let u = a^ x-1
and
v = 1/x
v = x^ -1

du/dx = a^x lna
and
dv/dx = -x^-2
= -1/x²

Now using formula:
dy/dx = U∙dv/dx + V∙du/dx
dy/dx = (a^ x-1)(-1/x²) + (x^ -1)(a^x lna)
dy/dx = -(a^ x-1) /x² + (a^x lna) /x
dy/dx = (1- a^ x) /x² + (a^x lna) /x

When x approaches zero, the answer will become undefined, but may be considered to be zero. (When you divide something by zero, it appears to approach infinity, but this has not been proven).

Consider:
n ÷ 0 = ∞ (or) 0 ??????????
0 ÷ 0 = 1 (or) 0 ??????????
0 ÷ n = 0

Answer 4

(a^(x-1))/x or ((a^x)-1)/x???

Either way as x approches 0, the function approaches infinity.

or taking the latter of the two options above, the numerator also approaches 0 and so the function would approach 0, infinity and 1 because 0/0 is 0, infinity and 1.

Anything divided by 0 is infinity
0 divided by anything is 0
anything divided by itself is 1

Answer 5

Since e^x = 1+x+x^2/2 + x^3/3 +x^4/4+..................i
and a^x= 1+log a (x) +{log a }^2(x)^2/2 +...................ii
Subtract 1 from both side
and a^x - 1= {log a} (x) +{log a }^2(x)^2/2 +...................
Divide both side by x, we get
{a^x-1}/x= {log a} +{log a }^2(x)/2 +...................
Now take the limits when x tends to 0
LHS = log a Ans

Answer 6

My response is : lim(...) = ln(a)..
Indeed : a^x = 1 +x.lin(a)+ 0.5*x^2.ln^2(a)+...
So, a^x-1=x.ln(a)+....
and (a^x-1)/x= ln(a) +x.(....)..

Answer 7

Anything divided by ZERO is INFINITY, not zero!
Get real!

Answer 8

lim (a^x -1))/x=lna it's basic stuff :D

Answer 9

a^x * ln(a) =

a^0 * ln(a) =

1 * ln(a) =

ln(a) =

Answer 10

ANYTHING DEVIDED INTO ZERO IS EQUAL TO ZERO.