Question

If $\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{1}{{{x^3}}}} \right)\left( {\dfrac{1}{{\sqrt {1 + x} }} - \dfrac{{1 + ax}}{{1 + bx}}} \right)$ exist, find the values of $a,\;b$. How can I solve this?

Answer 1

Hint : We can calculate that for $\mathop {\lim }\limits_{x \to 0}$ the given expression becomes $\dfrac{0}{0}$. Since this is an indeterminate form, we have to simplify the expression such that we can find a finite limit of the expression for $x \to 0$. We can use rationalizing of the denominator and then using binomial expansion for the term $\dfrac{1}{{\sqrt {1 + x} }}$ to simplify the expression.

Complete step by step solution:
We have been given that the limit $\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{1}{{{x^3}}}} \right)\left( {\dfrac{1}{{\sqrt {1 + x} }} - \dfrac{{1 + ax}}{{1 + bx}}} \right)$ exists. The value of this limit is not given. If we put $x = 0$ in the expression we will get,
$\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{1}{{{x^3}}}} \right)\left( {\dfrac{1}{{\sqrt {1 + x} }} - \dfrac{{1 + ax}}{{1 + bx}}} \right) = \left( {\dfrac{1}{0}} \right)\left( {\dfrac{1}{{\sqrt {1 + 0} }} - \dfrac{{1 + 0}}{{1 + 0}}} \right) = \left( {\dfrac{1}{0}} \right)\left( {1 - 1} \right) = \dfrac{0}{0}$
Since $\dfrac{0}{0}$ is an indeterminate form, we have to simplify the expression to get a finite value of the limit.
First we rationalize the term $\dfrac{1}{{\sqrt {1 + x} }}$ by multiplying by $\sqrt {1 + x}$ in the numerator and the denominator.

$$\Rightarrow \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{1}{{{x^3}}}} \right)\left( {\dfrac{1}{{\sqrt {1 + x} }} - \dfrac{{1 + ax}}{{1 + bx}}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{1}{{{x^3}}}} \right)\left( {\left( {\dfrac{1}{{\sqrt {1 + x} }} \times \dfrac{{\sqrt {1 + x} }}{{\sqrt {1 + x} }}} \right) - \dfrac{{1 + ax}}{{1 + bx}}} \right) \\\ = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{1}{{{x^3}}}} \right)\left( {\dfrac{{\sqrt {1 + x} }}{{1 + x}} - \dfrac{{1 + ax}}{{1 + bx}}} \right) \\\$$

We can simplify further as,

$$\Rightarrow \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{1}{{{x^3}}}} \right)\left( {\dfrac{{\sqrt {1 + x} }}{{1 + x}} - \dfrac{{1 + ax}}{{1 + bx}}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{1}{{{x^3}}}} \right)\left( {\dfrac{{\left( {1 + bx} \right)\sqrt {1 + x} - \left( {1 + ax} \right)\left( {1 + x} \right)}}{{\left( {1 + x} \right)\left( {1 + bx} \right)}}} \right) \\\ = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\left( {1 + bx} \right){{\left( {1 + x} \right)}^{\dfrac{1}{2}}} - \left( {1 + ax + x + a{x^2}} \right)}}{{{x^3}\left( {1 + x + bx + b{x^2}} \right)}}} \right) \;$$

Now we can use the binomial expansion formula to expand ${\left( {1 + x} \right)^{\dfrac{1}{2}}}$. The binomial formula is given as,
${\left( {1 + x} \right)^n} = 1 + nx + \dfrac{{n\left( {n - 1} \right)}}{{2!}}{x^2} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{x^3} + \;...\;\infty$
Thus,
${\left( {1 + x} \right)^{\dfrac{1}{2}}} = 1 + \dfrac{1}{2}x + \dfrac{{\dfrac{1}{2}\left( {\dfrac{1}{2} - 1} \right)}}{{2!}}{x^2} + \dfrac{{\dfrac{1}{2}\left( {\dfrac{1}{2} - 1} \right)\left( {\dfrac{1}{2} - 2} \right)}}{{3!}}{x^3} + \;...\;\infty \\\ \Rightarrow {\left( {1 + x} \right)^{\dfrac{1}{2}}} = 1 + \dfrac{x}{2} - \dfrac{{{x^2}}}{8} + \dfrac{{{x^3}}}{{16}} + ... \;$
Using this value in the limit we get,

$$= \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\left( {1 + bx} \right)\left( {1 + \dfrac{x}{2} - \dfrac{{{x^2}}}{8} + \dfrac{{{x^3}}}{{16}} + ...} \right) - \left( {1 + ax + x + a{x^2}} \right)}}{{{x^3}\left( {1 + x + bx + b{x^2}} \right)}}} \right) \\\ = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\left( {1 + \dfrac{x}{2} - \dfrac{{{x^2}}}{8} + \dfrac{{{x^3}}}{{16}} + ...} \right) + \left( {bx + \dfrac{{b{x^2}}}{2} - \dfrac{{b{x^3}}}{8} + \dfrac{{b{x^4}}}{{16}} + ...} \right) - \left( {1 + ax + x + a{x^2}} \right)}}{{{x^3}\left( {1 + x + bx + b{x^2}} \right)}}} \right) \\\ = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\dfrac{{\left( {1 + \dfrac{x}{2} - \dfrac{{{x^2}}}{8} + \dfrac{{{x^3}}}{{16}} + ...} \right)}}{{{x^3}}} + \dfrac{{\left( {bx + \dfrac{{b{x^2}}}{2} - \dfrac{{b{x^3}}}{8} + \dfrac{{b{x^4}}}{{16}} + ...} \right)}}{{{x^3}}} - \dfrac{{\left( {ax + x + a{x^2}} \right)}}{{{x^3}}}}}{{\left( {1 + x + bx + b{x^2}} \right)}}} \right) \\\ = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\left( {\dfrac{1}{{{x^3}}} + \dfrac{1}{{2{x^2}}} - \dfrac{1}{{8x}} + \dfrac{1}{{16}} + ...} \right) + \left( {\dfrac{b}{{{x^2}}} + \dfrac{b}{{2x}} - \dfrac{b}{8} + \dfrac{{bx}}{{16}} + ...} \right) - \left( {\dfrac{1}{{{x^3}}} + \dfrac{a}{{{x^2}}} + \dfrac{1}{{{x^2}}} + \dfrac{a}{x}} \right)}}{{\left( {1 + x + bx + b{x^2}} \right)}}} \right) \\\$$

We can further simplify this as,

$$l = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\left( {\dfrac{1}{{{x^3}}} + \dfrac{1}{{2{x^2}}} - \dfrac{1}{{8x}} + \dfrac{1}{{16}} + ...} \right) + \left( {\dfrac{b}{{{x^2}}} + \dfrac{b}{{2x}} - \dfrac{b}{8} + \dfrac{{bx}}{{16}} + ...} \right) - \left( {\dfrac{1}{{{x^3}}} + \dfrac{a}{{{x^2}}} + \dfrac{1}{{{x^2}}} + \dfrac{a}{x}} \right)}}{{\left( {1 + x + bx + b{x^2}} \right)}}} \right) \\\ \Rightarrow l = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\dfrac{1}{{{x^2}}}\left( {\dfrac{1}{2} - a + b - 1} \right) + \dfrac{1}{x}\left( {\dfrac{b}{2} - \dfrac{1}{8} - a} \right) + \left( {\dfrac{1}{{16}} + ... - \dfrac{b}{8} + \dfrac{{bx}}{{16}} + ...} \right)}}{{\left( {1 + x + bx + b{x^2}} \right)}}} \right) \;$$

Now, in this simplified form if we put $x = 0$ we still get indeterminate terms in the numerator.
The coefficient of $\dfrac{1}{{{x^2}}}$ and $\dfrac{1}{x}$ has to be equal to zero for the limit to exist as a finite number.
Thus,
$\left( {\dfrac{1}{2} - a + b - 1} \right) = 0$ and $\left( {\dfrac{b}{2} - \dfrac{1}{8} - a} \right) = 0$

$$\left( {\dfrac{1}{2} - a + b - 1} \right) = 0 \\\ \Rightarrow b - a = \dfrac{1}{2} \;$$

And,

$$\left( {\dfrac{b}{2} - \dfrac{1}{8} - a} \right) = 0 \\\ \Rightarrow b - 2a = \dfrac{1}{4} \;$$

We can solve this as,

$$\left( {b - a} \right) - \left( {b - 2a} \right) = \dfrac{1}{2} - \dfrac{1}{4} \\\ \Rightarrow a = \dfrac{1}{4} \;$$

And,

$$b - a = \dfrac{1}{2} \\\ \Rightarrow b = \dfrac{1}{2} + a = \dfrac{1}{2} + \dfrac{1}{4} = \dfrac{3}{4} \\\$$

Hence the required value is $a = \dfrac{1}{4}$ and $b = \dfrac{3}{4}$ .
So, the correct answer is “ $a = \dfrac{1}{4}$ and $b = \dfrac{3}{4}$ ”.

Note : Since the limit that we were getting after putting $x = 0$ in the given expression was indeterminate, we simplified the expression such that the indeterminate form was eliminated. We can also use L’Hospital’s rule when the limit is in the indeterminate form.