Find lim as x to - infinity (x+a) ln (1+ a/x)

Find $\displaystyle\lim_{x\to-\infty}\bigl(x+a\bigr)\ln\left(1+\frac{a}{x}\right)$.

Step 1: Identify the indeterminate form

As $x\to-\infty$ the ratio $\dfrac{a}{x}\to0$, so $\ln\bigl(1+\tfrac{a}{x}\bigr)\to\ln1 = 0$, while $x+a\to-\infty$. The product is of the form $(-\infty)\cdot0$.

Step 2: Substitute $t = 1/x$

Let $t = \dfrac1x$, so that $x = \dfrac1t$ and $t\to0^{-}$ as $x\to-\infty$. Then

\[ x+a = \frac1t+a = \frac{1+at}{t}, \qquad \frac{a}{x} = at, \]

and the expression becomes

\[ \bigl(x+a\bigr)\ln\left(1+\frac{a}{x}\right) = \frac{1+at}{t}\cdot\ln\bigl(1+at\bigr) = \frac{\bigl(1+at\bigr)\ln\bigl(1+at\bigr)}{t}. \tag{1} \]

Step 3: Expand via the Maclaurin series

With $u = at\to0$, \[\ln\bigl(1+at\bigr) = at-\frac{a^{2}t^{2}}{2}+O\!\left(t^{3}\right), \tag{2}\]

and multiplying by the prefactor $1+at$: \[\bigl(1+at\bigr)\ln\bigl(1+at\bigr) = at-\frac{a^{2}t^{2}}{2}+a^{2}t^{2}+O\!\left(t^{3}\right) = at+\frac{a^{2}t^{2}}{2}+O\!\left(t^{3}\right). \tag{3}\]

Step 4: Form the ratio and evaluate

\[ \frac{\bigl(1+at\bigr)\ln\bigl(1+at\bigr)}{t} = a+\frac{a^{2}t}{2}+O\!\left(t^{2}\right) \longrightarrow a \quad\text{as } t\to0. \]

Step 5: Verification

\[a = 2:\quad x = -100:\ 1.9798653; \qquad x = -1000:\ 1.9979987; \qquad x = -10^{5}:\ 1.9999800 \;\;\checkmark\]

\[\text{The correction } \frac{a^{2}}{2x} \text{ predicts } 2-0.002 = 1.998 \text{ at } x = -1000 \;\;\checkmark\]

\[\text{The same value } a \text{ is obtained as } x\to+\infty,\ \text{since only } t\to0 \text{ matters} \;\;\checkmark\]