Aspire Faculty ID #11994 · Topic: NIMCET 2025 · Just now
NIMCET 2025

What is the value of $\lim _{{x}\rightarrow\infty}-(x+1)\Bigg{(}{e}^{\frac{1}{x+1}}-1\Bigg{)}$?

Solution

We want to evaluate $ \displaystyle \lim_{x \to \infty} -(x+1)\left(e^{\frac{1}{x+1}} - 1\right). $ 

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 $.
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