NIMCET 2025 — Mathematics PYQ

Let the given limit be $L$:

$$L=\underset{x\to \infty }{\lim }-(x+1)\left(e^{\frac{1}{x+1}}-1\right)$$

Step 1: Identify the Indeterminate Form

As $x\to \infty$:

• $(x+1)\to \infty$

• $\frac{1}{x+1}\to 0$

• $\left(e^{\frac{1}{x+1}}-1\right)\to (e^0-1)=0$

This creates an indeterminate form of type $\infty \times 0$.

Step 2: Substitution Method

To simplify the limit, let us introduce a new variable $t$. Let:

$$t=\frac{1}{x+1}$$

Now, find the new limit for $t$ as $x$ approaches infinity:

As $x\to \infty$, the denominator $(x+1)\to \infty$, which means:

$$t\to 0$$

Also, from our substitution, we can rewrite $(x+1)$ as:

$$(x+1)=\frac{1}{t}$$

Step 3: Rewrite the Limit Equation

Substitute $t=\frac{1}{x+1}$ and $(x+1)=\frac{1}{t}$ back into the original limit expression:

$$L=\underset{t\to 0}{\lim }-\left(\frac{1}{t}\right)\left(e^t-1\right)$$

Rearranging the terms gives:

$$L=-\underset{t\to 0}{\lim }\frac{e^t-1}{t}$$

Step 4: Evaluate using the Standard Limits Identity

We know the standard fundamental limit theorem for exponential functions states that:

$$\underset{t\to 0}{\lim }\frac{e^t-1}{t}=1$$

Substituting this value back into our equation:

$$L=-(1)$$

$$L=-1$$

Final Answer

The correct option is (B).

$$\underset{x\to \infty }{\lim }-(x+1)\left(e^{\frac{1}{x+1}}-1\right)=-1$$