Rewrite it as a product: $ -(x+1)\left(e^{\frac{1}{x+1}} - 1\right) = -\dfrac{e^{\frac{1}{x+1}} - 1}{\frac{1}{x+1}}. $ 

 Now let $ t = \frac{1}{x+1} \Rightarrow t \to 0^+ $ as $ x \to \infty $. 

The expression becomes: $ -\dfrac{e^{t} - 1}{t}. $ 

Now apply L'Hospital’s Rule to the limit: $ \displaystyle \lim_{t \to 0} \dfrac{e^{t} - 1}{t} = \lim_{t \to 0} \dfrac{e^{t}}{1} = 1. $ 

 So the original limit is: $ -1. $ 

 Final Answer: $ -1 $.