Find lim as x to infinity (sqrt(1+x²ln ex/(x+1))-x)

Find $\displaystyle\lim_{x\to\infty}\left(\sqrt{1+x^{2}\ln\frac{ex}{x+1}}-x\right)$.

Step 1: Simplify the argument of the logarithm

\[ \ln\frac{ex}{x+1} = \ln e+\ln x-\ln\bigl(x+1\bigr) = 1-\ln\frac{x+1}{x} = 1-\ln\left(1+\frac1x\right). \tag{1} \]

Step 2: Expand the logarithm for large $x$

With $u = \dfrac1x\to0^{+}$ the Maclaurin series gives

\[ \ln(1+u) = u-\frac{u^{2}}{2}+\frac{u^{3}}{3}-\cdots = \frac1x-\frac{1}{2x^{2}}+O\!\left(x^{-3}\right), \tag{2} \]

and substituting (2) into (1), \[\ln\frac{ex}{x+1} = 1-\frac1x+\frac{1}{2x^{2}}+O\!\left(x^{-3}\right). \tag{3}\]

Step 3: Expand the radicand

Multiplying (3) by $x^{2}$ and adding $1$: \[x^{2}\ln\frac{ex}{x+1} = x^{2}-x+\frac12+O\!\left(x^{-1}\right), \tag{4}\]

\[ 1+x^{2}\ln\frac{ex}{x+1} = x^{2}-x+\frac32+O\!\left(x^{-1}\right). \tag{5} \]

Step 4: Expand the square root

Factor $x^{2}$ out of (5) and use $\sqrt{1+\varepsilon} = 1+\dfrac{\varepsilon}{2}-\dfrac{\varepsilon^{2}}{8}+\cdots$ with $\varepsilon = -\dfrac1x+\dfrac{3}{2x^{2}}+\cdots$:

\[ \sqrt{1+\varepsilon} = 1+\frac12\left(-\frac1x+\frac{3}{2x^{2}}\right)-\frac18\left(-\frac1x\right)^{2}+O\!\left(x^{-3}\right) = 1-\frac{1}{2x}+\frac{5}{8x^{2}}+O\!\left(x^{-3}\right), \]

where the $\varepsilon^{2}$ term contributes $-\dfrac{1}{8x^{2}}$ to the coefficient $\dfrac34-\dfrac18 = \dfrac58$. Multiplying by $x$, \[\sqrt{1+x^{2}\ln\frac{ex}{x+1}} = x-\frac12+\frac{5}{8x}+O\!\left(x^{-2}\right). \tag{6}\]

Step 5: Subtract $x$ and take the limit

\[ \sqrt{1+x^{2}\ln\frac{ex}{x+1}}-x = -\frac12+\frac{5}{8x}+O\!\left(x^{-2}\right) \longrightarrow -\frac12. \]

Step 6: Verification

\[x = 10:\ -0.4360582; \qquad x = 100:\ -0.4937354; \qquad x = 1000:\ -0.4993749 \;\;\checkmark\]

\[\text{The correction } \frac{5}{8x} \text{ predicts } -0.5+0.00625 = -0.49375 \text{ at } x = 100 \;\;\checkmark\]

\[\text{The factor } e \text{ inside the logarithm is essential: without it (5) would be } -x^{2}\ln\bigl(1+\tfrac1x\bigr) \;\;\checkmark\]