We want to evaluate \[ \lim_{x \to \infty} -(x+1)\left(e^{\tfrac{1}{x+1}} - 1\right). \] Step 1. Substitution. Let \(t = \tfrac{1}{x+1}\). Then as \(x \to \infty\), we have \(t \to 0^+\). The expression becomes \[ -(x+1)\left(e^{\tfrac{1}{x+1}} - 1\right) = -\frac{1}{t}\,(e^t - 1). \] Step 2. Known limit. We know that \[ \lim_{t \to 0} \frac{e^t - 1}{t} = 1. \] Hence, \[ \lim_{t \to 0^+} -\frac{1}{t}(e^t - 1) = -1. \] Final Answer: \[ \boxed{-1} \]