Question

JEE mains previous year Question .
Chapter:- limit , continuity and diffrentiabili​

Answer 1

EXPLANATION.

$\implies \displaystyle \lim_{x \to \infty} \bigg(\dfrac{x - 3}{x + 2} \bigg)^{x}$

As we know that,

Take x common from numerator and denominator of the equation, we get.

$\implies \displaystyle \lim_{x \to \infty} \dfrac{\bigg(1 - \dfrac{3}{x} \bigg)^{x} }{\bigg(1 + \dfrac{2}{x} \bigg)^{x} }$

$\implies \displaystyle \lim_{x \to \infty} = 1^{\infty}$

As we can see that,

It is  the indeterminant form of 1^(∞).

$\implies \displaystyle L = e^{\displaystyle \lim_{x \to \infty} \bigg(f(x) - 1 \bigg) \times g(x)}$

Using this formula in the equation, we get.

$\implies \displaystyle e^{ \displaystyle \lim_{x \to \infty}\bigg( \dfrac{x - 3}{x + 2} - 1 \bigg) (x)}$

$\implies \displaystyle e^{ \displaystyle \lim_{x \to \infty}\bigg( \dfrac{x - 3 - (x + 2)}{x + 2}\bigg) (x) }$

$\implies \displaystyle e^{\displaystyle \lim_{x \to \infty}\bigg( \dfrac{x - 3 - x - 2}{x + 2}\bigg)(x) }$

$\implies \displaystyle e^{ \displaystyle \lim_{x \to \infty}\bigg( \dfrac{-5}{x + 2} \bigg)(x)}$

Divide x in the denominator, we get.

$\implies \displaystyle e^{\displaystyle \lim_{x \to \infty}\bigg( \dfrac{-5}{(x + 2)/x} \bigg)}$

$\implies \displaystyle e^{ \displaystyle \lim_{x \to \infty}\bigg( \dfrac{- 5}{1 + (2/x)}\bigg) }$

$\implies \displaystyle e^{-5}$

Option [C] is correct answer.

Answer 2

$\huge{ \underline{ \color{teal}{ \textsf{ \textbf{Answer \: \: \: \: \: \: \: \: }}}}}$

let's write it in its form

$⇛ \sf \: lim \: x→∞ \: \: \: \: \: \: ( \frac{ \frac{x - 3}{x} }{ \frac{x + 2}{x} } )$

Divide numerator and denominator by X

$⇛ \sf \: lim \: x→∞ \: \: \: \: \: \: (\frac{ \frac{1 - 3}{x} }{ \frac{1 + 2}{x} } )$

$⇛ \sf 1$

Let's begin with solving the question

$⇛ \sf \: lim \: x→∞ \: \: \: \: \: \: 1 + f(x) {}^{2(x)}$

$\sf \: where \: lim \: f(x) = 0 \: and \: ∞→ \alpha$

$⇛ \sf \: lim \: \infty → \alpha \: \: \: \: \: \:e {}^{lim \: x - \alpha \: \: f(x) g(x)}$

$⇛ \sf \: lim \: x → \alpha \: \: \: \: \: \:( \frac{x - 3}{x + 2} ) {}^{x}$

$⇛ \sf \: lim \: x → \alpha \: \: \: \: \: \:( \frac{1 + x - 3}{x + 2} - 1) {}^{x}$

$⇛ \sf \: lim \: x → \alpha \: \: \: \: \: \:(1 + \frac{x - 3 - x - 2}{x + 2} ) {}^{x}$

$⇛ \sf \: lim \: x → \alpha \: \: \: \: \: \:(1 + \frac{ - 5}{x + 2} ) {}^{x}$

$\sf⇛lim \: e {}^{x - \alpha } \frac{ - 5}{x + 2} \times x$

$\sf⇛lim \: e {}^{x → \alpha } \: \: \: \frac{ - 5}{1 + 2 \div x}$

$⇛ \sf {e {}^{ - 5} }$

$\bold \red{Celestial} \bold{Centrix}$