Chapter 4, Problem 9P

To determine

The value of the sum of the constants $a+b+c+d$ in the equation.

Explanation of Solution

Given information:

The equation is,

  ${\lim }_{x\to 0}\frac{a{x}^2+\sin bx+\sin cx+\sin dx}{3x^2+5x^4+7x^6}=8$

Calculations:

Here the equation is,

  ${\lim }_{x\to 0}\frac{a{x}^2+\sin bx+\sin cx+\sin dx}{3x^2+5x^4+7x^6}=8$

Now the left side of the equation is of the form $\frac{0}{0}$ , therefore we can apply L’Hospotal’s rule.

Thus.

  $\begin{array}{l}{\lim }_{x\to 0}\frac{a{x}^2+\sin bx+\sin cx+\sin dx}{3x^2+5x^4+7x^6}\\ {\lim }_{x\to 0}\frac{2ax+b\cos bx+c\cos cx+d\cos dx}{6x+20x^3+42x^5}\mathrm{............}(1)\end{array}$

Again the equation is of the form $\frac{0}{0}$ , therefore we can apply L’Hospotal’s rule,

Thus.

  $\begin{array}{l}{\lim }_{x\to 0}\frac{2ax+b\cos bx+c\cos cx+d\cos dx}{6x+20x^3+42x^5}\\ {\lim }_{x\to 0}\frac{2a-b^2\sin bx-c^2\sin cx-d^2\sin dx}{6+60x^2+210x^4}\\ {\lim }_{x\to 0}\frac{2a}{6}\\ {\lim }_{x\to 0}\frac{a}{3}\end{array}$

Now,

  $\begin{array}{l}{\lim }_{x\to 0}\frac{a}{3}=8\\ a=8\times 3=24\end{array}$

Now the denominator of the equation is zero since $x\to 0$ . But the limits exists in the equation. So by equating the numerator equal to zero, we get.

, $\begin{array}{l}0+b+c+d=0\\ or,\\ b+c+d=0\end{array}$

Thus the value of $a+b+c+d=24.$