Find lim as x to -a (x+a) ln (1+ a/x), where a ≠0

Find $\displaystyle\lim_{x\to-a}\bigl(x+a\bigr)\ln\left(1+\frac{a}{x}\right)$, where $a\neq0$.

Step 1: Identify the form, and the side of approach

Note first that

\[ 1+\frac{a}{x} = \frac{x+a}{x}, \]

so as $x\to-a$ the argument of the logarithm tends to $0$ and $\ln\bigl(1+\tfrac{a}{x}\bigr)\to-\infty$, while the factor $x+a$ tends to $0$. The form is $0\cdot(-\infty)$.

The quotient $\dfrac{x+a}{x}$ must stay positive for the logarithm to be defined, and near $x = -a$ the denominator has the sign of $-a$. Hence only one side is available: $x\to-a^{-}$ when $a\gt 0$, and $x\to-a^{+}$ when $a\lt 0$. The limit below is understood on that side.

Step 2: Substitute $t = x+a$

With $t = x+a$, so that $x = t-a$ and $t\to0$, \[1+\frac{a}{x} = \frac{t}{t-a}, \qquad \bigl(x+a\bigr)\ln\left(1+\frac{a}{x}\right) = t\ln\frac{t}{t-a}. \tag{1}\]

Step 3: Split the logarithm

Since the ratio is positive while $t$ and $t-a$ may individually be negative, the split must be written with absolute values: \[t\ln\frac{t}{t-a} = t\ln|t| - t\ln|t-a|. \tag{2}\]

Step 4: Evaluate both terms

The first term is Problem 391 in disguise: $t\ln|t|\to0$ as $t\to0$, the power beating the logarithm. The second is harmless, since $|t-a|\to|a|\neq0$: $t\ln|t| \longrightarrow 0, \qquad t\ln|t-a| \longrightarrow 0\cdot\ln|a| = 0$.

Therefore the difference (2) tends to $0$.

Step 5: The same by l'Hôpital's rule

Writing (2) as a quotient and differentiating, \[\frac{\ln|t|-\ln|t-a|}{1/t} \;\longrightarrow\; \frac{\dfrac1t-\dfrac{1}{t-a}}{-\dfrac{1}{t^{2}}} = \bigl(-t^{2}\bigr)\cdot\frac{-a}{t\bigl(t-a\bigr)} = \frac{at}{t-a} \longrightarrow \frac{0}{-a} = 0.\]

Step 6: Verification

\[a = -1\ (x\to1^{+}):\quad x = 1.1:\ -0.2397895; \qquad x = 1.01:\ -0.0461512; \qquad x = 1.001:\ -0.0069088 \;\;\checkmark\]

\[a = 2\ (x\to-2^{-}):\quad x = -2.01:\ 0.0530330; \qquad x = -2.001:\ 0.0076014 \;\;\checkmark\]

\[\text{The decay is } t\ln|t|,\ \text{so the values shrink like } \tfrac{1}{100}\ln100 \text{ rather than like } t \;\;\checkmark\]