Prove that the limit of function f(x) is ln5, if function f(x) is given by f(x) =(5^x-5)/(x-1), x->0.

(5^0-5)/(0-1)

(1-5)/-1

4

I
am not getting ln5 as the limit of x as it approaches zero, as appears to have already
been mentioned, though for x -> 1 after testing for an indeterminant and applying
L'Hospital's rule

lim x->c f(x)/g(x) = lim
x->c f'(x)/g'(x)

f(x) =
5^x-5

f'(x) = 5^x ln(5)

g(x) =
x-1

g'(x) = 1

lim x->1
5^x
ln(5)/1

5ln(5)

The
major pitfall here is that L'Hospital's rule only applys
when

lim x->c f(x) = g(x) =
0

applied improperly, and let me state again,
this is improper and entirely wrong you would simplify the function
to

lim x->0 5^x
ln(5)/1

which would come out to
ln(5)

L'Hospital's rule works because an infintecimal is
comparable to another infintecimal.  But it does not change the fact that
 infintecimals become completely inconsequential when compared to any value that is not
itself an infintecimal.  If you think about it for a bit and about other instances in
mathematics like how 3x/2x = 3/2 or how only the largest power is considered in
fractions with limits to infinity it should become plainly obvious on what principles
this function is relying.