JAMIA 2023 — Mathematics PYQ

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Question

JAMIA 2023 — Mathematics PYQ

JAMIA | Mathematics | 2023

For $x \in R$ , $lim_{x \to \infty} (\frac{x - 3}{x + 2})^{x} =$

Choose the correct answer:

Explanation

Solution

1. General Formula for $1^\infty$ form:

For a limit of the form $\lim_{x \to a} [f(x)]^{g(x)}$ , if $f(x) \to 1$ and $g(x) \to \infty$ , the limit is given by:

e^{\lim_{x \to a} g(x) [f(x) - 1]}

2. Identify $f(x)$ and $g(x)$ :

In our case:

$f(x) = \dfrac{x-3}{x+2}$
$g(x) = x$

3. Apply the formula:

The limit becomes:

e^{\lim_{x \to \infty} x \left[ \dfrac{x-3}{x+2} - 1 \right]}

4. Simplify the expression inside the exponent:

Simplify the term $\left[ \dfrac{x-3}{x+2} - 1 \right]$ :

\frac{x-3 - (x+2)}{x+2} = \frac{x-3-x-2}{x+2} = \frac{-5}{x+2}

Now substitute this back into the limit:

e^{\lim_{x \to \infty} x \left( \frac{-5}{x+2} \right)}

e^{\lim_{x \to \infty} \frac{-5x}{x+2}}

5. Evaluate the limit of the exponent:

Divide the numerator and denominator by $x$ :

\lim_{x \to \infty} \frac{-5}{1 + \frac{2}{x}} = \frac{-5}{1 + 0} = -5

Final Answer:

Substituting the result back into the exponential:

e^{-5}