How do we evaluate limit of,
lim x-> 0 [ln(x+1)/( (2^x) - 1)]
I tried using the substitution x+1 = e^k , when x tends to 0 so does k, which gave out,
lim k->0 [ k/((2^((e^k) - 1)) -1 ) ]
which I simplified into( for the ease of use let e^k =a)
lim k->0 [k/( ((2^a)/2) - 1)]
lim k->0 [2k/((2^a) -2 )]
which sums up to,
(2*0)/((2^(e^0)) - 2)
(2*0)/(2-2)
which gives out an undefined value as 0/0
How do we solve this?
How about good old L'Hopital's rule
lim x-> 0 [ln(x+1)/( (2^x) - 1)]
= lim x-> 0 [1/(x+1) / (ln2*2^x)
= (1/1)/(ln2 ( 1)
= 1/ln2 = appr 1.44269..
check: let x = .0001 in the original
= ln(1.0001) / (2^.0001 - 1)
= 1.4425 , not bad, my answer is correct
To evaluate the limit of the given function, we can use a different approach. Let's start from the original expression:
lim x->0 [ln(x + 1) / ((2^x) - 1)]
Since we have an indeterminate form of 0/0, we can apply L'Hôpital's rule, which states that if we have a limit of the form 0/0 or ∞/∞, and the derivative of the numerator and denominator both exist, then the limit is equal to the limit of the derivatives of the numerator and denominator.
Let's find the derivatives of the numerator and denominator:
d/dx[ln(x + 1)] = 1 / (x + 1)
d/dx[(2^x) - 1] = ln(2) * 2^x
Now, we can rewrite the limit using the derivatives:
lim x->0 [ (1 / (x + 1)) / (ln(2) * 2^x) ]
Now, as x approaches 0, the denominator (ln(2) * 2^x) approaches ln(2).
Therefore, the limit becomes:
(1 / (ln(2)))
So, the value of the limit is 1 / ln(2).