Without using L'Hospital Rule, evaluate $\lim_{x\to 1} \frac{x}{\ln x}-\frac{1}{x \ln x}$.

Question

Given Limit: $L=\lim_{x\to 1} \frac{x}{\ln x}-\frac{1}{x \ln x}$
The limit reduces to $L=2$ via L'Hospital Rule
Even after employing the logarithmic limit $\lim_{x\to 0}\frac{\ln (1+x)}{x}=1$
Also after logarithmic series expansion $\ln (1+x) =x-\frac{x^2}{x}+\frac{x^3}{3}-\frac{x^4}{4}+...$
Couldn't attain the answer Please explain possible algebraic manipulation or procedure to figure out the solution.

Answer 1

HINT:

Note that we have

$$\frac{x}{\log(x)}-\frac1{x\log(x)}=\frac{x+1}{x}\frac{x-1}{\log(x)}$$

and

$$\frac{x-1}{x}\le \log(x)\le x-1$$

which can be obtained using the limit definition of the exponential function along with Bernoulli's inequality as I showed in THIS ANSWER.

Answer 2

Hint: set $y+1=x$ then $L=\lim_\limits{y\to 0} \frac{y+1}{\ln(y+1)}-\frac{1}{(y+1) \ln(y+1)}=\lim_\limits{y\to 0} \frac{y}{\ln(y+1)}\cdot\frac{y+2}{y+1}$

Answer 3

One can split terms into addition by applying logarithm again:

$$L = \lim_{x \to 1}{\frac{\frac{x^2 - 1}{x}}{\ln x}} = \lim_{x \to 1}\frac{\big(1 + \frac{1}{x}\big) \cdot \big(1 - \frac{1}{x}\big)}{\ln x}$$

Taking natural log:

$$\ln{L} = \lim_{x \to 1}{\ln{\big(1 + \frac{1}{x}\big)}} + \lim_{x \to 1}{\ln\big[{\frac{\big(1 + \frac{1}{x}\big)}{\ln{x}}}}\big]$$

The rest is an exercise.

Answer 4

We have $${x\over \ln x}-{1\over x\ln x}={x^2-1\over x\ln x}={2\over x}{x^2-1\over \ln(x^2)-\ln 1}$$ The limit is equal ${2\over \ln'(1)}=2.$