Here,is a limit problem: $\lim \limits _{x \to 0} {x^3 \over {\sqrt {a+x}} (bx – \sin x)} = 1$. Here, $a \in \mathbb R _+$. The question is to find the values of $a$ and $b$.
Here is my workout. Dividing the numerator and denominator by $x$, $\lim \limits _{x \to 0} {x^2 \over {b \sqrt {a+x} – \sqrt {a+x}}} = 1$. Rationalising and simplifying, $\lim \limits _{x \to 0} {x^2 \over {\sqrt a (b-1)}} = 1$.
I could not do beyond this, please anyone help.
Best Answer
http://www.wolframalpha.com/input/?i=taylor%28sqrt%28a%2Bx%29%28bx-sin%28x%29%29%29
shows the taylor expansion of $\sqrt{a+x}(bx-sin(x)$
Choose $a$ and $b$ in such a way, that the coefficients at $x$ and $x^2$ become $0$ and the coefficient at $x^3$ becomes $1$. This gives $a=36$ and $b=1$
http://www.wolframalpha.com/input/?i=limit+x^3%2F%28sqrt%2836%2Bx%29%28x-sin%28x%29%29%29+%2C+x+tends+to+0
shows that the limit is indeed $1$