<s>[INST] Limit Problem-To find the value of a and b,when the value of the limit is given. 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.
 [/INST] Suppose that $b \neq 1$.
Concerning $a$, there are two possibilities. Assume first that $a \neq 0$. Then $\sqrt {a+x} \to \sqrt a$. The rest of your expression becomes ${x^3 \over {bx - \sin x}} = {x^2 \over {b - {\sin x \over x}}}$, which will tend to $0$, so your whole limit will be $0$, not $1$. If $a=0$, then your expresion becomes ${x^{3 \over 2} \over {b - {\sin x \over x}}}$, which again tends to $0$.
This shows that necessarily $b=1$.
Assume now that $a=0$. Your expression becomes ${x^{5 \over 2} \over {x - \sin x}}$; apply l'Hopital's theorem twice, reducing it to ${5 \over 2} {x^{3 \over 2} \over {1 - \cos x}}$, then to ${15 \over 4} {x^{1 \over 2} \over {\sin x}}$, which tends to $\infty$ (also, because of the $1 \over 2$ in the power, the previous expression is defined only for $x>0$). We conclude that $a \neq 0$.
Finally, $\lim \limits _{x \to 0} {x^3 \over {\sqrt {a+x}} (x - \sin x)} = {1 \over {\sqrt a}} \lim \limits _{x \to 0} {x^3 \over {x - \sin x}} = {3 \over {\sqrt a}} \lim \limits _{x \to 0} {x^2 \over {1 - \cos x}} = {6 \over {\sqrt a}} \lim \limits _{x \to 0} {x \over {\sin x}} = {6 \over {\sqrt a}}$. Since this must be $1$, necessarily $a=36$.
To conclude: $a=36, \space b=1$.
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