# 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.

• Where did the sin-term go ? May 30 '15 at 8:25
• You could try the rule of L'hospital or use the taylor expansion for $\sqrt{a+x}(bx-sin(x))$ May 30 '15 at 8:26
• On dividing by x,lim x tends to 0 sinx/x=1 May 30 '15 at 8:27

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 $1$

• I agree with your answer, I've deleted mine. +1! May 30 '15 at 8:55
• Please avoid comments for chatting. MSE offers a dedicated chat section for this kind of conversations. May 30 '15 at 9:02
• In the first one, only "a part" of all the terms was "hyperlinked". Thanks. May 30 '15 at 9:03

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: $a=36, \space b=1$.

• Correct and complete rigorous answer. +1 May 30 '15 at 9:40

To show the limit is 1 it suffices to note that \begin{equation*} \frac{x^{3}}{\sqrt{36+x}(x-\sin x)}=\frac{1}{\sqrt{36+x}\left( \frac{x-\sin x% }{x^{3}}\right) } \end{equation*} and \begin{eqnarray*} \lim_{x\rightarrow 0}\sqrt{36+x} &=&6 \\ \lim_{x\rightarrow 0}\frac{x-\sin x}{x^{3}} &=&\lim_{x\rightarrow 0}\frac{% 1-\cos x}{3x^{2}}=\lim_{x\rightarrow 0}\frac{\sin x}{6x}=\frac{1}{6}. \end{eqnarray*}

• And, of course, one might wonder how you got to the numbers $36$ and $1$, which was the whole point of the problem. May 30 '15 at 9:33
• @Alex M. Peter ends his solution by : show that the limit is $\color{blue}{ indeed}$ 1. It is that what I did only. May 30 '15 at 9:38