Question

If the $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{1}{{{x}^{3}}}\left( \dfrac{1}{\sqrt{1+x}}-\dfrac{1+ax}{1+bx} \right)$ exists and has the value equal to l, then find the value of $\dfrac{1}{a}-\dfrac{2}{l}+\dfrac{3}{b}$.

Answer 1

Expand ${{\left( 1+x \right)}^{\dfrac{1}{2}}}$ by using the binomial expansion of any series of any index. Use the condition given that limit exists and finite to find a, b, and l.

Complete step by step answer:
We have an expression $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{1}{{{x}^{3}}}\left( \dfrac{1}{\sqrt{1+x}}-\dfrac{1+ax}{1+bx} \right)$ which exists and has a value equal to l, then we need to determine $\dfrac{1}{a}-\dfrac{2}{l}+\dfrac{3}{b}=?$
We have,
$l=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{1}{{{x}^{3}}}\left( \dfrac{1}{\sqrt{1+x}}-\dfrac{1+ax}{1+bx} \right)....\left( i \right)$
Let us directly put the limits to the expression given, we get
$l=\dfrac{1}{0}\left( \dfrac{1}{\sqrt{1+0}}-\dfrac{1+0}{1+0} \right)$
$l=\dfrac{1}{0}\left( 1-1 \right)=\dfrac{0}{0}$
Hence, the given limit expression is in indeterminate form. Hence, we need to simplify the given relation before proceeding for putting limits.
Let us simplify the term $\dfrac{1}{\sqrt{1+x}}$ by rationalizing it. Multiply by $\sqrt{1+x}$ to numerator and denominator to $\dfrac{1}{\sqrt{1+x}}$ in expression, we get,
$l=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{1}{{{x}^{3}}}\left( \dfrac{1}{\sqrt{1+x}}\times \dfrac{\sqrt{1+x}}{\sqrt{1+x}}-\dfrac{1+ax}{1+bx} \right)$
$l=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{1}{{{x}^{3}}}\left( \dfrac{\sqrt{1+x}}{1+x}-\dfrac{\left( 1+ax \right)}{1+bx} \right)$
We can write the expression as:
$l=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left( 1+bx \right)\sqrt{1+x}-\left( 1+ax \right)\left( 1+x \right)}{{{x}^{3}}\left( 1+x \right)\left( 1+bx \right)}$
$l=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left( 1+bx \right){{\left( 1+x \right)}^{\dfrac{1}{2}}}-\left( 1+ax+x+a{{x}^{2}} \right)}{{{x}^{3}}\left( 1+bx+x+b{{x}^{2}} \right)}$
$l=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left( 1+bx \right){{\left( 1+x \right)}^{\dfrac{1}{2}}}-\left( 1+ax+x+a{{x}^{2}} \right)}{{{x}^{3}}+b{{x}^{4}}+{{x}^{4}}+b{{x}^{5}}}....\left( ii \right)$
We know the expansion of ${{\left( 1+x \right)}^{n}}$ for any index n can be given as
${{\left( 1+x \right)}^{n}}=1+nx+\dfrac{n\left( n-1 \right)}{2!}{{x}^{2}}+\dfrac{n\left( n-1 \right)\left( n-2 \right)}{3!}{{x}^{3}}+....\left( iii \right)$
We can expand ${{\left( 1+x \right)}^{\dfrac{1}{2}}}$ by using the above relation, we will get
${{\left( 1+x \right)}^{\dfrac{1}{2}}}=1+\dfrac{1}{2}x+\dfrac{\left( \dfrac{1}{2} \right)\left( \dfrac{1}{2}-1 \right)}{2!}{{x}^{2}}+\dfrac{\left( \dfrac{1}{2} \right)\left( \dfrac{1}{2}-1 \right)\left( \dfrac{1}{2}-2 \right)}{3!}{{x}^{3}}+.....$
Or,
${{\left( 1+x \right)}^{\dfrac{1}{2}}}=1+\dfrac{x}{2}-\dfrac{1}{8}{{x}^{2}}+\dfrac{1}{16}{{x}^{3}}+.....\left( iv \right)$
Putting the value of ${{\left( 1+x \right)}^{\dfrac{1}{2}}}$ as expression in equation (iv) in equation (ii), we get
$l=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left( 1+bx \right)\left( 1+\dfrac{x}{2}-\dfrac{{{x}^{2}}}{8}+\dfrac{{{x}^{3}}}{16}+.... \right)-\left( 1+ax+x+a{{x}^{2}} \right)}{{{x}^{3}}+b{{x}^{4}}+{{x}^{4}}+{{x}^{5}}}$
On simplifying the above relation, we get
$l=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left( 1+\dfrac{x}{2}-\dfrac{{{x}^{2}}}{8}+\dfrac{{{x}^{3}}}{16} \right)+\left( bx+\dfrac{b{{x}^{2}}}{2}-\dfrac{b{{x}^{3}}}{8}+\dfrac{b{{x}^{4}}}{16} \right)-\left( 1+ax+x+a{{x}^{2}} \right)}{{{x}^{3}}+b{{x}^{4}}+{{x}^{4}}+{{x}^{5}}}$
Taking the same coefficients in one bracket, we get
$l=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x\left( \dfrac{1}{2}+b-a-1 \right)+{{x}^{2}}\left( \dfrac{-1}{8}+\dfrac{b}{2}-a \right)+{{x}^{3}}\left( \dfrac{1}{16}-\dfrac{b}{8} \right)+\dfrac{b{{x}^{4}}}{16}}{{{x}^{3}}+b{{x}^{4}}+{{x}^{4}}+{{x}^{5}}}$
Taking out x from numerator and denominator, we get
$l=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left( b-a-\dfrac{1}{2} \right)+x\left( -\dfrac{1}{8}+\dfrac{b}{2}-a \right)+{{x}^{2}}\left( \dfrac{1}{16}-\dfrac{b}{8} \right)+\dfrac{b{{x}^{3}}}{16}}{{{x}^{2}}+b{{x}^{3}}+{{x}^{3}}+{{x}^{4}}}$
Dividing the whole equation by ${{x}^{2}}$, we get
$l=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{\left( b-a-\dfrac{1}{2} \right)}{{{x}^{2}}}+\dfrac{\left( \dfrac{-1}{8}+\dfrac{b}{2}-a \right)}{x}+\left( \dfrac{1}{16}-\dfrac{b}{8} \right)+\dfrac{xb}{16}}{1+bx+x+{{x}^{2}}}....\left( v \right)$
Now, if we put a limit $x\to 0$ to the given expression, we get positive infinity or negative infinity if $\left( b-a-\dfrac{1}{2} \right)$ and $\left( \dfrac{-1}{8}+\dfrac{b}{2}-a \right)$ will be any definite value except 0. But it is given that ‘l’ is a finite value of a given limit. Hence, $\left( b-a-\dfrac{1}{2} \right)$ and $\left( \dfrac{-1}{8}+\dfrac{b}{2}-a \right)$ should be zero and hence ‘l’ can be written as
$l=\dfrac{\left( \dfrac{1}{16}-\dfrac{b}{8} \right)+0}{1+0}$
$l=\dfrac{1}{16}-\dfrac{b}{8}....\left( vi \right)$
And,
$b-a-\dfrac{1}{2}=0....\left( vii \right)$
$\dfrac{-1}{8}+\dfrac{b}{2}-a=0....\left( viii \right)$
Let us solve the equation (vii) and (viii) to get values of ‘a’ and ‘b’.
From equation (vii), we have
$a=b-\dfrac{1}{2}$
Putting the value of ‘a’ in equation (viii),
We get,
$\dfrac{-1}{8}+\dfrac{b}{2}-\left( b-\dfrac{1}{2} \right)=0$
$\dfrac{-1}{8}+\dfrac{b}{2}-b+\dfrac{1}{2}=0$
$\dfrac{-b}{2}=\dfrac{-1}{2}+\dfrac{1}{8}=\dfrac{-4+1}{8}=\dfrac{-3}{8}$
$b=\dfrac{3}{4}$
And hence, $a=\dfrac{3}{4}-\dfrac{1}{2}=\dfrac{3-2}{4}=\dfrac{1}{4}$
$a=\dfrac{1}{4}$
Now, we can calculate ‘l’ by equation (vi), as
$l=\dfrac{1}{16}-\dfrac{3}{32}=\dfrac{2-3}{32}$
$l=\dfrac{-1}{32}$
Hence, now we have
$a=\dfrac{1}{4},b=\dfrac{3}{4}$ and $l=\dfrac{-1}{32}....\left( ix \right)$
Now coming to question, we have to find $\dfrac{1}{a}-\dfrac{2}{l}+\dfrac{3}{b}=?$
Putting values of a, l, b from equation (ix), we get
$4+64+4=72$

So, the correct answer is “72”.

Note: Another approach for this question would be that one can apply L’Hospital Rule to the given expression as
$l=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left( 1+bx \right)-\left( 1+ax \right)\sqrt{1+x}}{{{x}^{3}}\left( 1+bx \right)\sqrt{1+x}}$
Differentiating numerator and denominator with respect to x, we get
$l=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{b-\left( a\sqrt{1+x}+\dfrac{\left( 1+ax \right)}{2}{{\left( 1+x \right)}^{\dfrac{1}{2}}} \right)}{\left( 3{{x}^{2}}+4b{{x}^{3}} \right)\sqrt{1+x}+\dfrac{{{x}^{3}}+b{{x}^{4}}}{2}{{\left( 1+x \right)}^{\dfrac{-1}{2}}}}$
On simplifying the given relation, we get
$l=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{2b\sqrt{1+x}-\left( 2a+3ax+1 \right)}{2\left( 1+x \right)\left( 3{{x}^{2}}+4b{{x}^{3}} \right)+\left( {{x}^{3}}+b{{x}^{4}} \right)}$
Now, putting $x\to 0$, we get
$l=\dfrac{2b-\left( 2a+1 \right)}{0}$
As l is finite, hence 2b – 2a – 1 = 0.
And we need to apply to L'Hospital to ‘l’ again. Now, we get
$l=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{2b}{2}\dfrac{1}{\sqrt{1+x}}-3a}{2\left( 3{{x}^{2}}+4b{{x}^{3}} \right)+2\left( 1+x \right)\left( 6x+12b{{x}^{2}} \right)+3{{x}^{2}}+4b{{x}^{3}}}$
Now, putting $x\to 0$, we get
$l=\dfrac{b-3a}{0}$
As l is finite, hence b – 3a = 0 or b = 3a.
Now, calculate a and b by using the above both equations and apply L’Hospital Rule again to ‘l’ to get the value of the limit. The calculation is the most important part of the question as well. So, we need to take care of the calculation as well. One can think that why higher powers of x are ignored in the solution. The reason is simple for it that limit to higher powers of x will be zero, i.e. if ${{x}^{3}}$ is tending to zero, then why we write ${{x}^{4}}$ or any other high power.