Key Concepts and Formulas

• Taylor Series Expansions: For small values of $x$, we can approximate functions using their Taylor series expansions around $x=0$. Key expansions include:
  • $e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \ldots$
  • $\cos u = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \ldots$
• Limit Properties: For a limit of a fraction to exist and be finite, if the denominator approaches zero, the numerator must also approach zero, and the order of the terms in the numerator must be of at least the same order as the terms in the denominator.
• Trigonometric Identities: The double-angle formula for cosine, $\cos(2x) = 1 - 2\sin^2(x)$, is useful for simplifying expressions involving $\cos(2x)$.

Step-by-Step Solution

Step 1: Expand the numerator and denominator using Taylor Series. We are given the limit: $\lim_{x \rightarrow 0} \frac{e^{a x}-\cos (b x)-\frac{cx e^{-c x}}{2}}{1-\cos (2 x)} = 17$ We will use Taylor series expansions around $x=0$ for the functions involved.

For the numerator:

• $e^{ax} = 1 + ax + \frac{(ax)^2}{2!} + O(x^3) = 1 + ax + \frac{a^2x^2}{2} + O(x^3)$
• $\cos(bx) = 1 - \frac{(bx)^2}{2!} + O(x^4) = 1 - \frac{b^2x^2}{2} + O(x^4)$
• $e^{-cx} = 1 - cx + \frac{(-cx)^2}{2!} + O(x^3) = 1 - cx + \frac{c^2x^2}{2} + O(x^3)$ So, $\frac{cx e^{-cx}}{2} = \frac{cx}{2} \left(1 - cx + \frac{c^2x^2}{2} + O(x^3)\right) = \frac{cx}{2} - \frac{c^2x^2}{2} + \frac{c^3x^3}{4} + O(x^4)$.

Substituting these into the numerator: $N(x) = \left(1 + ax + \frac{a^2x^2}{2}\right) - \left(1 - \frac{b^2x^2}{2}\right) - \left(\frac{cx}{2} - \frac{c^2x^2}{2}\right) + O(x^3)$ $N(x) = (1-1) + (a - \frac{c}{2})x + \left(\frac{a^2}{2} + \frac{b^2}{2} + \frac{c^2}{2}\right)x^2 + O(x^3)$ $N(x) = \left(a - \frac{c}{2}\right)x + \frac{a^2 + b^2 + c^2}{2}x^2 + O(x^3)$

For the denominator:

• $1 - \cos(2x)$. Using the identity $\cos(2x) = 1 - 2\sin^2(x)$, we get $1 - \cos(2x) = 2\sin^2(x)$.
• For small $x$, $\sin(x) = x - \frac{x^3}{3!} + O(x^5)$.
• So, $\sin^2(x) = \left(x - \frac{x^3}{6} + O(x^5)\right)^2 = x^2 - 2x \frac{x^3}{6} + O(x^6) = x^2 - \frac{x^4}{3} + O(x^6)$.
• Therefore, $D(x) = 1 - \cos(2x) = 2\sin^2(x) = 2\left(x^2 - \frac{x^4}{3} + O(x^6)\right) = 2x^2 - \frac{2x^4}{3} + O(x^6)$. The dominant term in the denominator is $2x^2$.

Step 2: Analyze the limit based on the dominant terms. The limit becomes: $\lim_{x \rightarrow 0} \frac{\left(a - \frac{c}{2}\right)x + \frac{a^2 + b^2 + c^2}{2}x^2 + O(x^3)}{2x^2 + O(x^4)} = 17$ For this limit to exist and be finite (equal to 17), the terms in the numerator of lower order than $x^2$ must cancel out. This means the coefficient of the $x$ term in the numerator must be zero.

Step 3: Equate the coefficient of $x$ in the numerator to zero. $a - \frac{c}{2} = 0$ This implies: $c = 2a$

Step 4: Rewrite the limit with the simplified numerator and denominator. With $a - \frac{c}{2} = 0$, the numerator simplifies to: $N(x) = \frac{a^2 + b^2 + c^2}{2}x^2 + O(x^3)$ The limit now is: $\lim_{x \rightarrow 0} \frac{\frac{a^2 + b^2 + c^2}{2}x^2 + O(x^3)}{2x^2 + O(x^4)} = 17$ Dividing both numerator and denominator by $x^2$: $\lim_{x \rightarrow 0} \frac{\frac{a^2 + b^2 + c^2}{2} + O(x)}{2 + O(x^2)} = 17$ As $x \rightarrow 0$, the limit is: $\frac{\frac{a^2 + b^2 + c^2}{2}}{2} = 17$

Step 5: Solve for the expression involving $a$, $b$, and $c$. $\frac{a^2 + b^2 + c^2}{4} = 17$ $a^2 + b^2 + c^2 = 17 \times 4$ $a^2 + b^2 + c^2 = 68$

Step 6: Substitute the relationship between $a$ and $c$ to find the required expression. We found that $c = 2a$. Substitute this into the equation from Step 5: $a^2 + b^2 + (2a)^2 = 68$ $a^2 + b^2 + 4a^2 = 68$ $5a^2 + b^2 = 68$

Common Mistakes & Tips

• Incomplete Taylor Expansions: Ensure you expand the Taylor series to a sufficient order. For this problem, expanding up to the $x^2$ term in the numerator and $x^2$ term in the denominator was necessary to determine the limit.
• Algebraic Errors: Be meticulous with algebraic manipulations, especially when dealing with fractions and squares of terms.
• Denominator Simplification: Correctly simplifying the denominator $1 - \cos(2x)$ using trigonometric identities or Taylor series is crucial. Using $1-\cos(2x) = 2\sin^2(x)$ and then $\sin(x) \approx x$ for small $x$ leads to $1-\cos(2x) \approx 2x^2$.

Summary

The problem requires evaluating a limit that results in an indeterminate form. By using Taylor series expansions for the exponential and cosine functions in the numerator and the denominator, we can express the limit in terms of powers of $x$. For the limit to be a finite non-zero value, the coefficient of the lowest power of $x$ in the numerator must be zero. This condition allows us to establish a relationship between $a$ and $c$. Subsequently, by considering the ratio of the coefficients of the next higher power of $x$ (which is $x^2$ in this case), we can form an equation involving $a^2$, $b^2$, and $c^2$. Substituting the relationship between $a$ and $c$ into this equation yields the desired expression $5a^2 + b^2$.

The final answer is $\boxed{68}$.