Finding the Limit of ln(x)/(x-1) as x Approaches 1 Without L'H?pital's Rule

To find the limit

[lim_{xto 1}frac{ln x}{x - 1}]

without using L'H?pital's rule, we can use the Taylor series expansion of (ln x) around (x 1).

The Taylor Series Expansion Method

The Taylor series expansion of (ln x) around (x 1) is:

(ln x x - 1 - frac{(x - 1)^2}{2} - frac{(x - 1)^3}{3} - cdots)

For values of (x) close to 1, we can approximate (ln x) by only considering the first term of the series:

(ln x approx x - 1)

Now, substituting this approximation into our limit gives:

[lim_{xto 1}frac{ln x}{x - 1} approx lim_{xto 1}frac{x - 1}{x - 1}]

As long as (x eq 1), this simplifies to:

(1)

Thus, the limit is:

[boxed{1}]

Defining the Derivative Method

By definition of the derivative, we have:

[f'(a) lim_{h to 0}frac{f(a h) - f(a)}{h}]

Letting (a h x) when (h to 0) implies (a to x), we get:

[f'(a) lim_{x to a}frac{f(x) - f(a)}{x - a}]

Replaced (a) by (1), we get:

[f'(1) lim_{xto 1}frac{f(x) - f(1)}{x - 1}]

Let (f(x) ln x), so (f'(x) frac{1}{x}), and (f(1) ln 1 0). Then, we have:

[f'(1) lim_{xto 1}frac{ln x - ln 1}{x - 1} 1]

Hence, the limit is:

[boxed{1}]

Why L'H?pital's Rule is Not Suitable

It is a logical fallacy to use L'H?pital's rule to find the limit of the quotient (frac{ln x}{x - 1}) as (x to 1) since one of the conditions to apply L'H?pital's rule is to know the derivative of the function at the specific point. In this case, to find the derivative of (ln x) at (x 1), you need to know the limit of (frac{ln x - ln 1}{x - 1}). You would end up in a logical loop if you were to apply L'H?pital's rule here.

Using the Exponential Function

Let (y ln x), which means (x e^y). Since (ln 1 0), we have (e^0 1). Now, let (t ln x), so (x e^t) and (x to 1) is (t to 0).

Consider the properties of the exponential function (e^t): for every (t) and (s),

(e^{t s} e^t e^s) (e^s e^{-s} e^{s - s} e^0 1) Since the exponential function is positive, if we multiply both sides of the above by (e^t), we get:

[e^t e^{-t} 1]

Therefore, for all (t in -1, 1), we have:

[1 leq frac{e^t e^{-t}}{2t} leq frac{1}{1 - |t|}]

Dividing by (t) once for (0

[1 leq frac{e^t e^{-t}}{2t^2} leq frac{1}{1 - |t|}]

Applying the squeeze rule, we obtain:

[lim_{t to 0} frac{e^t e^{-t}}{2t^2} 1]

Consequently, we find:

[lim_{t to 0} frac{e^t - e^{-t}}{2t} 1]

Since (ln x t), we get:

[lim_{x to 1} frac{ln x - ln 1}{x - 1} 1]

Hence, the limit is:

[boxed{1}]