Evaluate L = lim as x to infinity x ln x {1- [(ln
Evaluate $\displaystyle L = \lim_{x\to\infty}x\ln x\left\{1-\left[\frac{\ln\bigl(x-1\bigr)}{\ln x}\right]^{a}\right\}$.
Explanation of Method
A bracket of size 1/(x ln x), multiplied by exactly x ln x.
Step 1: Expand the base
Write $\ln(x-1) = \ln x+\ln\bigl(1-\tfrac1x\bigr)$, so that
\[ \frac{\ln\bigl(x-1\bigr)}{\ln x} = 1+\frac{\ln\bigl(1-\tfrac1x\bigr)}{\ln x}. \]
Using $\ln(1-u) = -u-\dfrac{u^{2}}{2}-\cdots$ with $u = \dfrac1x$: $\ln\left(1-\frac1x\right) = -\frac1x-\frac{1}{2x^{2}}+O\!\left(x^{-3}\right)$.
\[ \frac{\ln\bigl(x-1\bigr)}{\ln x} = 1-\frac{1}{x\ln x}-\frac{1}{2x^{2}\ln x}+O\!\left(\frac{1}{x^{3}\ln x}\right). \tag{1} \]
Step 2: Raise the base to the power $a$
Put $\delta(x) = \dfrac{\ln(x-1)}{\ln x}-1 = -\dfrac{1}{x\ln x}+O\!\left(\dfrac{1}{x^{2}\ln x}\right)$, so that
\[ \left[\frac{\ln\bigl(x-1\bigr)}{\ln x}\right]^{a} = \bigl(1+\delta\bigr)^{a} = \exp\Bigl(a\ln\bigl(1+\delta\bigr)\Bigr). \]
Since $\ln(1+\delta) = \delta-\dfrac{\delta^{2}}{2}+O(\delta^{3})$ and $\delta^{2} = O\!\left(\dfrac{1}{x^{2}\ln^{2}x}\right)$ is of lower order, \[a\ln\bigl(1+\delta\bigr) = -\frac{a}{x\ln x}+O\!\left(\frac{1}{x^{2}\ln x}\right). \tag{2}\]
Step 3: Expand the exponential
The exponent in (2) tends to $0$, so $e^{s} = 1+s+O(s^{2})$ gives
\[ \left[\frac{\ln\bigl(x-1\bigr)}{\ln x}\right]^{a} = 1-\frac{a}{x\ln x}+O\!\left(\frac{1}{x^{2}\ln x}\right). \tag{3} \]
Step 4: Form the bracket and multiply
\[ 1-\left[\frac{\ln\bigl(x-1\bigr)}{\ln x}\right]^{a} = \frac{a}{x\ln x}+O\!\left(\frac{1}{x^{2}\ln x}\right), \tag{4} \]
\[ x\ln x\left\{1-\left[\frac{\ln\bigl(x-1\bigr)}{\ln x}\right]^{a}\right\} = a+O\!\left(\frac1x\right). \tag{5} \]
Step 5: Take the limit
The remainder in (5) vanishes, so
\[ L = \lim_{x\to\infty}\left[a+O\!\left(\frac1x\right)\right] = a. \]
Step 6: Verification
\[a = 2:\quad x = 100:\ 2.0078738; \qquad x = 10^{4}:\ 2.0000891; \qquad x = 10^{6}:\ 2.0000009 \;\;\checkmark\]
\[\text{Dominant asymptotics: } \left(1-\frac{1}{x\ln x}\right)^{a}\approx1-\frac{a}{x\ln x} \;\Longrightarrow\; x\ln x\cdot\frac{a}{x\ln x} = a \;\;\checkmark\]
\[a = 1 \text{ reduces the bracket to } \frac{\ln x-\ln(x-1)}{\ln x},\ \text{giving } x\ln\frac{x}{x-1}\to1 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,L = a\,}\]
Remark
Remark. The factor $x\ln x$ is chosen to match the bracket exactly — any other normalisation would give $0$ or $\infty$. Note also how mildly the exponent $a$ enters: to first order, raising $1+\delta$ to the power $a$ merely multiplies $\delta$ by $a$, which is why the answer is $a$ itself and not something like $a^{2}$ or $2^{a}$. The convergence is slow because the neglected terms carry an extra $\ln x$ only in the denominator.