lim x→0 (ax²eˣ - b logₑ(1+x) + cxe⁻ˣ) / (x² sin x) = 1 then 16(a² + b² + c²) is equal to _______.

Q: Why can't we just use L'Hôpital's Rule here? It seems like it would be faster.

L'Hôpital's Rule is a powerful tool, but applying it multiple times to a complex expression involving products like $x^2 e^x$ and $x^2 \sin x$ often leads to algebraic nightmares. The Maclaurin series method is much cleaner and more systematic for limits involving $x \to 0$.

Q: Why do we set the coefficients of $x$ and $x^2$ to zero?

Think about the limit $\lim_{x \to 0} \frac{P(x)}{x^3}$. If $P(x)$ has terms like $Ax + Bx^2$, the limit would be $\lim_{x \to 0} (\frac{A}{x^2} + \frac{B}{x} + C)$, which blows up to infinity. To keep the limit finite and non-zero, those lower-power terms must vanish.

Q: How do I know how many terms of the Maclaurin series to expand?

Always match the power of the denominator. Since our denominator simplifies to $x^3$, we need to expand the numerator up to the $x^3$ term. Any higher-order terms (like $x^4$) will vanish when divided by $x^3$ as $x \to 0$.

Question

$\lim_{x 	o 0} \frac{ax^2e^x - b\log_e(1+x) + cxe^{-x}}{x^2 \sin x} = 1$ then $16(a^2 + b^2 + c^2)$ is equal to _______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with finding the constants $a, b,$ and $c$ such that the following limit holds:

$$\lim_{x 	o 0} \frac{ax^2e^x - b\ln(1+x) + cxe^{-x}}{x^2 \sin x} = 1$$

As $x 	o 0$, we know that $\sin x \approx x$. Therefore, the denominator behaves as $x^2(x) = x^3$. For the limit to be a finite non-zero value, the numerator must also behave as $k x^3$ near $x=0$.

### The Maclaurin Arsenal

To analyze the numerator, we expand the functions into their Maclaurin series:

$$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^4)$$

$$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} + O(x^4)$$

$$e^{-x} = 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + O(x^4)$$

Substituting these into the numerator $N(x) = ax^2e^x - b\ln(1+x) + cxe^{-x}$:

$$N(x) = ax^2(1 + x + \frac{x^2}{2} + \dots) - b(x - \frac{x^2}{2} + \frac{x^3}{3} + \dots) + cx(1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \dots)$$

### The Balancing Act

Grouping the terms by powers of $x$, we obtain:

$$N(x) = (c - b)x + (a + \frac{b}{2} - c)x^2 + (\frac{a}{2} - \frac{b}{3} + \frac{c}{2})x^3 + O(x^4)$$

For the limit to exist and equal $1$, the coefficients of $x$ and $x^2$ must be zero:

1) $c - b = 0 \Rightarrow c = b$

2) $a + \frac{b}{2} - c = 0 \Rightarrow a + \frac{b}{2} - b = 0 \Rightarrow a = \frac{b}{2} \Rightarrow b = 2a$

Substituting $c=b$ and $b=2a$ into the coefficient of $x^3$:

$$\frac{a}{2} - \frac{2a}{3} + \frac{2a}{2} = 1$$

$$\frac{3a - 4a + 6a}{6} = 1 \Rightarrow \frac{5a}{6} = 1 \Rightarrow a = \frac{6}{5}$$

Wait, re-evaluating the coefficient of $x^3$: The term is $(\frac{a}{2} - \frac{b}{3} + \frac{c}{2})x^3$. With $b=2a$ and $c=2a$:

$$\frac{a}{2} - \frac{2a}{3} + \frac{2a}{2} = \frac{3a - 4a + 6a}{6} = \frac{5a}{6} = 1 \Rightarrow a = \frac{6}{5}$$

### Final Calculation

Given the values $a = \frac{6}{5}$, $b = \frac{12}{5}$, and $c = \frac{12}{5}$, we calculate the final expression:

$$16(a^2 + b^2 + c^2) = 16 \left( \left(\frac{6}{5}ight)^2 + \left(\frac{12}{5}ight)^2 + \left(\frac{12}{5}ight)^2 ight)$$

$$16 \left( \frac{36 + 144 + 144}{25} ight) = 16 \left( \frac{324}{25} ight) = \frac{5184}{25} = 207.36$$

**The final result is 207.36.**

Animated Solution for Mathematics - Limits, Continuity and Differentiability: $lim_{x \to 0} \frac{a x ^{2} e ^{x} - b l o g _{e} ( 1 + x ) + c x e ^{- x}}{x ^{2} s i n x} = 1$ then $16 (a^{2} + b^{2} + c^{2})$ is equal to _______.

Enter Numerical Value:

Visualized Solution

$lim_{x \to 0} \frac{a x ^{2} e ^{x} - b l o g _{e} ( 1 + x ) + c x e ^{- x}}{x ^{2} s i n x} = 1$

Denominator Approximation

Maclaurin Series Expansions

Numerator Substitution

Grouping Coefficients

Eliminating Lower Powers

Evaluating the Limit

Calculating $b$ and $c$

Final Computation

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

Analyzing the Setup

The Maclaurin Arsenal

The Balancing Act

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; } limx→0​x2sinxax2ex−bloge​(1+x)+cxe−x​=1 then 16(a2+b2+c2) is equal to _______.