If lim x→0 (e^(ax) - cos(bx) - (cxe^(-cx))/2) / (1 - cos(2x)) = 17, then 5a² + b² is equal to

Q: Wait, why do we stop at the $x^2$ term? Why not go to $x^3$ or $x^4$?

That is a brilliant question! In limits, we are looking for the 'dominant' behavior as $x$ approaches zero. Our denominator, $1 - \cos(2x)$, behaves like $2x^2$. If we had higher-order terms like $x^3$ or $x^4$ in the numerator, they would vanish when divided by $x^2$ as $x \to 0$. Conversely, if we stopped too early (say, at the constant term), we would miss the information needed to calculate the limit. We match the order of the numerator to the order of the denominator.

Q: Why must the coefficient of $x$ be zero? What if it isn't?

If the coefficient of the $x$ term in the numerator were non-zero, say it was $K$, then our limit would look like $\lim_{x \to 0} \frac{Kx + \dots}{2x^2}$. This simplifies to $\lim_{x \to 0} (\frac{K}{2x} + \dots)$, which blows up to infinity! Since the problem states the limit is a finite value (17), the universe of this problem forces that $x$ term to vanish. It is a mathematical necessity for existence.

Q: Can I use L'Hopital's Rule instead of Taylor Series?

You absolutely can, but be warned: L'Hopital's Rule requires you to differentiate the numerator and denominator twice. With the exponential and trigonometric functions involved, the chain rule will make the algebra very messy and prone to calculation errors. Taylor Series is the 'sniper' approach—it is cleaner, faster, and much less likely to lead to a sign error.

Question

If $\lim_{x 	o 0} \frac{e^{ax} - \cos(bx) - \frac{cxe^{-cx}}{2}}{1 - \cos(2x)} = 17$, then $5a^2 + b^2$ is equal to

Options:

(A) 64

(B) 72

(C) 68

(D) 76

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with evaluating the limit:
$$\lim_{x 	o 0} \frac{e^{ax} - \cos(bx) - \frac{cxe^{-cx}}{2}}{1 - \cos(2x)} = 17$$
Our goal is to determine the value of $5a^2 + b^2$.

### The Denominator's Secret

Every limit problem has a 'ruler'—the term that defines the scale of the problem. Here, it is the denominator: $1 - \cos(2x)$.

Using the trigonometric identity $1 - \cos(2x) = 2\sin^2(x)$, we observe that as $x 	o 0$, $\sin(x) \approx x$. Therefore, the denominator behaves as $2x^2$.

This implies that we only need to consider terms in the numerator up to the order of $x^2$. Any terms of higher order will vanish as $x 	o 0$.

### The Taylor Series Toolkit

To handle the transcendental functions, we utilize Taylor series expansions near $x=0$:
$$e^u \approx 1 + u + \frac{u^2}{2}$$
$$\cos(u) \approx 1 - \frac{u^2}{2}$$

Expanding our specific terms:
1. $e^{ax} \approx 1 + ax + \frac{a^2x^2}{2}$
2. $\cos(bx) \approx 1 - \frac{b^2x^2}{2}$
3. $\frac{cxe^{-cx}}{2} \approx \frac{cx(1 - cx)}{2} = \frac{cx}{2} - \frac{c^2x^2}{2}$

### The Numerator's Convergence

Assembling these expansions into the numerator, we obtain:
$$\left(1 + ax + \frac{a^2x^2}{2}ight) - \left(1 - \frac{b^2x^2}{2}ight) - \left(\frac{cx}{2} - \frac{c^2x^2}{2}ight)$$

The constant terms $1 - 1$ cancel out. Grouping the remaining terms by powers of $x$:
$$\left(a - \frac{c}{2}ight)x + \left(\frac{a^2 + b^2 + c^2}{2}ight)x^2$$

### The Constraint of Finiteness

We now examine the limit:
$$\lim_{x 	o 0} \frac{\left(a - \frac{c}{2}ight)x + \left(\frac{a^2 + b^2 + c^2}{2}ight)x^2}{2x^2} = 17$$

For the limit to be finite, the coefficient of the $x$ term must be zero. Thus, $a - \frac{c}{2} = 0$, which implies $c = 2a$.

With the $x$ term eliminated, the expression simplifies to:
$$\frac{\frac{a^2 + b^2 + c^2}{2}}{2} = 17$$
$$\Rightarrow a^2 + b^2 + c^2 = 68$$

### Final Calculation

Substituting $c = 2a$ into our equation:
$$a^2 + b^2 + (2a)^2 = 68$$
$$a^2 + b^2 + 4a^2 = 68$$
$$5a^2 + b^2 = 68$$

The final value is **68**.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $lim_{x \to 0} \frac{e ^{a x} - c o s ( b x ) - \frac{c x e ^{- c x}}{2}}{1 - c o s ( 2 x )} = 17$ , then $5 a^{2} + b^{2}$ is equal to

Select Answer:

Visualized Solution

The Given Limit

Simplifying the Denominator

Taylor Series Expansions

Expanding $e^{a x}$

Expanding $cos (b x)$

Expanding $\frac{c x e ^{- c x}}{2}$

Combining Numerator Terms

Condition for Finite Limit

Equating the Limit Value

Final Substitution

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

The Denominator's Secret

The Taylor Series Toolkit

The Numerator's Convergence

The Constraint of Finiteness

Final Calculation

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Animated Solution for Mathematics - Limits, Continuity and Differentiability: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If limx→0​1−cos(2x)eax−cos(bx)−2cxe−cx​​=17, then 5a2+b2 is equal to