Find lim as x to infinity [x-x²ln (1+ frac1x) ]

Find $\displaystyle\lim_{x\to\infty}\left[x-x^{2}\ln\left(1+\frac1x\right)\right]$.

Step 1: Identify the indeterminate form

As $x\to\infty$ we have $\ln\bigl(1+\tfrac1x\bigr)\to0^{+}$, and since $\ln\bigl(1+\tfrac1x\bigr)\sim\dfrac1x$ the second term behaves like $x^{2}\cdot\dfrac1x = x$. Both terms therefore grow without bound and the difference is of the form $\infty-\infty$.

Step 2: Substitute $t = 1/x$

With $t = \dfrac1x$, so that $x = \dfrac1t$ and $t\to0^{+}$, \[x-x^{2}\ln\left(1+\frac1x\right) = \frac1t-\frac{\ln(1+t)}{t^{2}}, \tag{1}\]

and combining over a common denominator, \[= \frac{t-\ln(1+t)}{t^{2}}. \tag{2}\]

At $t = 0$ both parts vanish: the form is $\dfrac00$.

Step 3: Expand via the Maclaurin series

\[ \ln(1+t) = t-\frac{t^{2}}{2}+\frac{t^{3}}{3}-\cdots \;\Longrightarrow\; t-\ln(1+t) = \frac{t^{2}}{2}-\frac{t^{3}}{3}+\cdots \tag{3} \]

Step 4: Form the ratio and evaluate

\[ \frac{t-\ln(1+t)}{t^{2}} = \frac{\dfrac{t^{2}}{2}-\dfrac{t^{3}}{3}+\cdots}{t^{2}} = \frac12-\frac{t}{3}+\cdots \longrightarrow \frac12 \quad\text{as } t\to0^{+}. \]

Step 5: Verification

\[x = 10:\ 0.4689820; \qquad x = 100:\ 0.4966915; \qquad x = 1000:\ 0.4996669 \;\;\checkmark\]

\[\text{In terms of } x \text{ the expansion reads } \frac12-\frac{1}{3x}+\cdots:\ \text{at } x = 100,\ 0.49667 \;\;\checkmark\]

\[\text{Equivalently } x^{2}\ln\left(1+\frac1x\right) = x-\frac12+\frac{1}{3x}-\cdots \;\;\checkmark\]