Question

If $L={lim}_{x\to 0}\frac{1}{x^3}(\frac{1}{\sqrt{1+x}}-\frac{1+ax}{1+bx})$ exists then

• A. $a=\frac{1}{4}$
• B. $b=\frac{3}{4}$
• C. $L=-\frac{1}{32}$
• D. $L=\frac{1}{32}$

Answer 1

The correct option is C $L=-\frac{1}{32}$

$L=\underset{x\to 0}{lim}\frac{(\frac{1}{\sqrt{1+x}}-\frac{1+ax}{1+bx})}{x^3}$

$\frac{0}{0}$ form $⟹$ aplly L-Hospital Rule

$L=\underset{x\to 0}{lim}\frac{-\frac{1}{2}(1+x{)}^{-\frac{3}{2}}-\frac{a-b}{(1+bx{)}^2}}{3x^2}=\frac{-\frac{1}{2}-(a-b)}{0}⟹a-b=-\frac{1}{2}(\because \frac{0}{0}form)$

$L=\underset{x\to 0}{lim}\frac{\frac{3}{4}(1+x{)}^{-\frac{5}{2}}-b(1+bx{)}^{-3}}{6x}=\frac{\frac{3}{4}-b}{0}⟹b=\frac{3}{4}⟹a=\frac{1}{4}$

$L=\underset{x\to 0}{lim}\frac{-\frac{15}{8}(1+x{)}^{-\frac{7}{2}}-\frac{9}{16}(1+\frac{3}{4}x{)}^{-4}}{6}=-\frac{1}{32}$