Let a,b,c,d be real numbers such that limx→0 sin dx+ cx^5 + bx^3 + ax/x^5 = 0. If b= 4/3 then what is the value of 15ac+ 2?

We expand \(\sin(d\cdot x)\) using its Taylor series at x = 0.

 

\(\begin{array}{rcl} \sin(d\cdot x) &=& \displaystyle\sum_{k = 0}^\infty \dfrac{(d\cdot x)^{2k+1}(-1)^k}{(2k + 1)!}\\ &=&\displaystyle\sum_{k = 0}^\infty \dfrac{d^{2k+1}(-1)^k}{(2k + 1)!} x^{2k+1}\\ &=&d\cdot x - \dfrac{d^3}{6}x^3+\dfrac{d^5}{120}x^5-\mathcal O (x^7) \end{array}\)

 

For \(\displaystyle\lim_{x\to0}\dfrac{\sin(d\cdot x) + cx^5 + bx^3 + ax}{x^5}\) to exist, the polynomial part must eliminate all terms with degrees lower than 5 of the Taylor expansion above.

Therefore, we have \(\begin{cases}a = -d\\b = \dfrac{d^3}6\\\end{cases}\). But we also know that \(b = \dfrac43\).

Comparing gives \(\dfrac{d^3}6 = \dfrac43\). Solving gives \(d = 2\).

 

Then since \(a = -d\), we have \(a = -2\).

 

Therefore, we have:

\(\begin{array}{rcl} \displaystyle\lim_{x\to0}\dfrac{\sin(d\cdot x) + cx^5 + bx^3 + ax}{x^5} &=& \displaystyle\lim_{x\to 0}\dfrac1{x^5}\left(d\cdot x - \dfrac{d^3}6x^3+\dfrac{d^5}{120}x^5-\mathcal O(x^7)+cx^5 + \dfrac{d^3}6x^3 - d\cdot x\right)\\ &=& \displaystyle\lim_{x\to0}\left(\dfrac{d^5}{120} + c - \mathcal O(x^2)\right)\\ &=& \displaystyle\lim_{x\to0}\left(\dfrac{2^5}{120} + c - \mathcal O(x^2)\right)\\ &=& \displaystyle\lim_{x\to0}\left(\dfrac{4}{15} + c - \mathcal O(x^2)\right)\\ &=& \dfrac{4}{15} + c \end{array}\)

 

But we also know that this limit is 0. Then:

\(\dfrac{4}{15} + c = 0\\ c = -\dfrac{4}{15} \)

 

Now we know the values of a and c, so 

\(\begin{array}{rcl} 15ac + 2 &=& 15 \cdot (-2) \cdot \dfrac{-4}{15} + 2\\ &=& 10 \end{array}\)

 

Remark: If you are not familiar with the Big \(\mathcal O\) notation, just treat it as \(\cdots\).