Question

$\underset{x\to \infty }{\lim }{\left(1-\frac{4}{\left(x-1\right)}\right)}^{3x-1}=$

• A. $e^{12}$
• B. $e^{-12}$
• C. $e^4$
• D. $e^3$

Answer 1

The correct option is B

$e^{-12}$

Explanation for the correct option:

Finding the value of $\underset{x\to \infty }{\lim }{\left(1-\frac{4}{\left(x-1\right)}\right)}^{3x-1}$:

The given equation is,

$$\begin{gathered} I=\underset{x\to \infty }{\lim }{\left(1-\frac{4}{\left(x-1\right)}\right)}^{3x-1} \\ \left[{\left(1-0\right)}^{\infty }\to 1^{\infty }\text{from}\right] \end{gathered}$$

We know that

$\underset{x\to a}{\lim }f{\left(x\right)}^{g\left(x\right)}=e^{\underset{x\to a}{\lim }g\left(x\right)\left(f\left(x\right)-1\right)}$

Then,

$$\begin{gathered} I=e^{\underset{x\to \infty }{\lim }\left(3x-1\right)\left(1-\frac{4}{\left(x-1\right)}-1\right)} \\ \Rightarrow I=e^{\underset{x\to \infty }{\lim }\left(3x-1\right)\left(\frac{-4}{\left(x-1\right)}\right)} \\ \Rightarrow I=e^{\underset{x\to \infty }{\lim }\left(-4\right)\left(\frac{x\left(3-{\displaystyle \frac{1}{x}}\right)}{x\left(1-{\displaystyle \frac{1}{x}}\right)}\right)} \\ \Rightarrow I=e^{\underset{x\to \infty }{\lim }\left(-4\right)\frac{\left(3-{\displaystyle \frac{1}{x}}\right)}{\left(1-{\displaystyle \frac{1}{x}}\right)}} \end{gathered}$$

At,

$$\begin{gathered} x\to \infty \\ \frac{1}{x}\to \frac{1}{\infty }\to 0 \end{gathered}$$

Then,

$$\begin{gathered} I=e^{\left(-4\right)\frac{\left(3-{\displaystyle 0}\right)}{\left(1-{\displaystyle 0}\right)}} \\ \Rightarrow I=e^{-4\times 3} \\ \Rightarrow I=e^{-12} \end{gathered}$$

Hence, the correct answer is option (B).