The value of the function given f(x) = (5^x-5)/(x-1), for x --> 0 can be arrived at by substituting the value of x = 0 in the expression for the function f(x) = (5^x-5)/(x-1)
=> (5^0 - 5)/( 0 - 1)
=> ( 1 - 5) / (0 - 1)
The required value of lim x-->0 [ (5^x-5)/(x-1)] = +4 not ln 5.
To apply remarcable limits, in this case, x approaches to 1 and not to 0.
We notice that if we'll substitute x by 1, we'll get an indetermination.
lim (5^x-5)/(x - 1) = (5^1 - 5)/(1-1) = (5 - 5)/0 = 0/0
We'll apply L'Hospital to determine the limit:
lim (5^x-5)/(x-1) = lim (5^x-5)'/(x-1)'
lim (5^x-5)'/(x-1)' = lim 5^x*ln 5/1 =5ln5
lim (5^x-5)/(x-1) = 5*ln 5
For x->1, the limit of the function is: lim (5^x-5)/(x-1) = ln 5^5.
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
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.