Question

If $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \dfrac{a}{x} - \dfrac{4}{{{x^2}}}} \right)^{2x}} = {e^3}$, then $a$ is equal to:
$a)\,\dfrac{2}{3} \\\ b)\,\dfrac{3}{2} \\\ c)\,2 \\\ d)\,\dfrac{1}{2} \\\$

Answer 1

We are given $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \dfrac{a}{x} - \dfrac{4}{{{x^2}}}} \right)^{2x}} = {e^3}$, we will firstly observe the intermediate form ${1^\infty }$in the given expression.Then using if $f(x) \to 1$for $x \to \infty$and $g(x) \to \infty$for $x \to \infty$
$\mathop {\lim }\limits_{x \to \infty } {\left( {f(x)} \right)^{g(x)}} = {e^{\mathop {\lim }\limits_{x \to \infty } \left( {f(x) - 1} \right)g(x)}}$.Using these concept we try to solve the question.

Complete step-by-step answer:
$\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \dfrac{a}{x} - \dfrac{4}{{{x^2}}}} \right)^{2x}} = {e^3}\,\,\,\,\,\,\,\,\, \to (1)$
Here firstly we try to solve L.H.S of (1)
Firstly,
We will consider $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \dfrac{a}{x} - \dfrac{4}{{{x^2}}}} \right)^{2x}}$and will try to solve this limit, we can observe that $\left( {1 + \dfrac{a}{x} - \dfrac{4}{{{x^2}}}} \right) \to 1$ for $x \to \infty$ and for $2x \to \infty$for $x \to \infty$.
So for $x \to \infty ,\,\,\,{\left( {1 + \dfrac{a}{x} - \dfrac{4}{{{x^2}}}} \right)^{2x}} \to {1^\infty }$
So we can say that ${1^\infty }$ is an intermediate form.
And we know if $f(x) \to 1$for $x \to \infty$and $g(x) \to \infty$for $x \to \infty$, then
$\mathop {\lim }\limits_{x \to \infty } {\left( {f(x)} \right)^{g(x)}} = {e^{\mathop {\lim }\limits_{x \to \infty } \left( {f(x) - 1} \right)g(x)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to (2)$
Now using (2) solve $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \dfrac{a}{x} - \dfrac{4}{{{x^2}}}} \right)^{2x}}$where we take $f(x) = \left( {1 + \dfrac{a}{x} - \dfrac{4}{{{x^2}}}} \right)$ and $g(x) = 2x$
So we get

$$\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \dfrac{a}{x} - \dfrac{4}{{{x^2}}}} \right)^{2x}} = {e^{\mathop {\lim }\limits_{x \to \infty } 2x\left( {\left( {1 + \dfrac{a}{x} - \dfrac{4}{{{x^2}}}} \right) - 1} \right)}} \\\ = {e^{\mathop {\lim }\limits_{x \to \infty } 2x\left( {\left( {\dfrac{a}{x} - \dfrac{4}{{{x^2}}}} \right)} \right)}} \\\ = {e^{\mathop {\lim }\limits_{x \to \infty } 2a - \dfrac{4}{x}}} \\\ \mathop {\lim }\limits_{x \to \infty } {\left( {1 + \dfrac{a}{x} - \dfrac{4}{{{x^2}}}} \right)^{2x}} = {e^{2a}}\,\,\,\,\,\,\,\,\,\, \to (3) \\\$$

Now from (1) we know that
$\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \dfrac{a}{x} - \dfrac{4}{{{x^2}}}} \right)^{2x}} = {e^3}$
Using (3) we get,
${e^3} = {e^{2a}}$
Now comparing powers we get,
$2a = 3 \\\ a = \dfrac{3}{2} \\\$

So, the correct answer is “Option B”.

Note: In this type of questions we always try to find out the intermediate form, if any. Also in (3), we used $\mathop {\lim }\limits_{x \to \infty } 2x\left( {\dfrac{a}{x} - \dfrac{4}{{{x^2}}}} \right) = 2a$ because we know $\mathop {\lim }\limits_{x \to \infty } \dfrac{1}{x} = 0$ and $\mathop {\lim }\limits_{x \to \infty } y = y$