If $\displaystyle\lim_{x \to 0} \frac{e^{(a-1)x}+2\cos bx+(c-2)e^{-x}}{x\cos x-\log_e(1+x)} = 2$, then $a^2+b^2+c^2$ is equal to :
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Solution Steps
1
Check the form of the limit
Denominator at $x=0$: $0 \cdot \cos 0 – \log_e(1+0) = 0 – 0 = 0$.
For the limit to be finite, the numerator must also be $0$ at $x=0$.
Numerator at $x=0$: $e^0 + 2\cos 0 + (c-2)e^0 = 1 + 2 + c – 2 = c+1$
So $c + 1 = 0 \Rightarrow \boxed{c = -1}$
2
Expand numerator using Taylor series (with c = −1)
$e^{(a-1)x} = 1 + (a-1)x + \dfrac{(a-1)^2}{2}x^2 + \cdots$
$2\cos bx = 2\!\left(1 – \dfrac{b^2x^2}{2} + \cdots\right) = 2 – b^2x^2 + \cdots$
$(c-2)e^{-x} = -3e^{-x} = -3\!\left(1 – x + \dfrac{x^2}{2} – \cdots\right) = -3 + 3x – \dfrac{3x^2}{2} + \cdots$
3
Collect terms by power of x in numerator
Constant: $(1+2-3) = 0$ ✓
Coefficient of $x$: $(a-1)+0+3 = a+2$
Coefficient of $x^2$: $\dfrac{(a-1)^2}{2} – b^2 – \dfrac{3}{2}$
So numerator $= (a+2)x + \left[\dfrac{(a-1)^2}{2} – b^2 – \dfrac{3}{2}\right]x^2 + \cdots$
4
Expand denominator using Taylor series
$x\cos x = x\!\left(1 – \dfrac{x^2}{2} + \cdots\right) = x – \dfrac{x^3}{2} + \cdots$
$\log_e(1+x) = x – \dfrac{x^2}{2} + \dfrac{x^3}{3} – \cdots$
$x\cos x – \log_e(1+x) = \dfrac{x^2}{2} + O(x^3)$
5
Condition: limit is finite ⇒ coefficient of x in numerator = 0
Since denominator starts at $x^2$, numerator must also start at $x^2$ for a finite limit.
$\Rightarrow a + 2 = 0 \Rightarrow \boxed{a = -2}$
Now the numerator becomes:
$\left[\dfrac{(-3)^2}{2} – b^2 – \dfrac{3}{2}\right]x^2 = \left[\dfrac{9}{2} – b^2 – \dfrac{3}{2}\right]x^2 = (3 – b^2)x^2$
6
Apply the limit condition = 2
$$\lim_{x \to 0}\frac{(3-b^2)x^2}{\frac{x^2}{2}} = 2(3-b^2) = 2$$
$3 – b^2 = 1 \Rightarrow \boxed{b^2 = 2}$
7
Compute a² + b² + c²
$a = -2,\quad b^2 = 2,\quad c = -1$
$$a^2+b^2+c^2 = (-2)^2 + 2 + (-1)^2 = 4+2+1 = \mathbf{7}$$
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Key Insight
Three unknowns → three conditions: (1) Num=0 at x=0 gives c; (2) coeff of x=0 gives a; (3) ratio of x² coefficients gives b².
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Theory
1. Taylor Series Expansion Method for Limits
When evaluating limits of the $\frac{0}{0}$ indeterminate form, the Taylor series method is the most systematic and powerful technique for JEE. Expand every function about $x = 0$ up to the required order. The key expansions to memorize are: $e^{kx} = 1 + kx + \frac{k^2x^2}{2} + \frac{k^3x^3}{6} + \cdots$; $\cos(bx) = 1 – \frac{b^2x^2}{2} + \frac{b^4x^4}{24} + \cdots$; $\sin(bx) = bx – \frac{b^3x^3}{6} + \cdots$; $\log(1+x) = x – \frac{x^2}{2} + \frac{x^3}{3} – \cdots$. After expanding, collect terms by powers of $x$. Cancel the lowest power common to both numerator and denominator. The limit equals the ratio of the leading surviving coefficients. This method avoids repeated application of L’Hôpital’s rule and is significantly faster in exam conditions.
2. Condition for Existence of Finite Limit
For $\lim_{x \to 0} \frac{f(x)}{g(x)}$ to be a finite nonzero value $L$: (i) Both $f(0) = 0$ and $g(0) = 0$ (indeterminate form). (ii) If $g(x) \sim Cx^n$ near $x=0$, then $f(x)$ must also start at $x^n$; if $f(x)$ starts at a lower power, the limit is $\pm\infty$; if at a higher power, limit is $0$. In this problem, the denominator $x\cos x – \log(1+x) \sim \frac{x^2}{2}$ (starts at $x^2$). So numerator must also start at $x^2$ for a finite nonzero limit. This yields two conditions on the constants: (1) constant term of numerator $= 0$ → fixes $c$; (2) coefficient of $x$ in numerator $= 0$ → fixes $a$. These matching conditions are the essence of parametric limit problems in JEE.
3. Denominator Expansion: x cos x − log(1+x)
This denominator appears frequently in JEE limits. Expanding carefully: $x\cos x = x\left(1 – \frac{x^2}{2} + \frac{x^4}{24} – \cdots\right) = x – \frac{x^3}{2} + \cdots$ and $\log_e(1+x) = x – \frac{x^2}{2} + \frac{x^3}{3} – \frac{x^4}{4} + \cdots$. Subtracting: $x\cos x – \log(1+x) = (x – \frac{x^3}{2}) – (x – \frac{x^2}{2} + \frac{x^3}{3}) + \cdots = \frac{x^2}{2} – \frac{x^3}{2} – \frac{x^3}{3} + \cdots = \frac{x^2}{2} – \frac{5x^3}{6} + \cdots$. The leading term is $\frac{x^2}{2}$. In problems where you need the next term’s coefficient for a second-order limit, include the $x^3$ term too. Recognizing this standard denominator saves time in exams.
4. Parametric Limit Problems — Strategy
When a limit problem involves unknown constants $(a, b, c, \ldots)$ and a given limit value $L$, follow this systematic strategy: Step 1 — Set up the indeterminate form and find how many unknowns you can determine from the existence condition. Step 2 — Match orders of $x$ in numerator and denominator to ensure the limit is finite. Each order-matching gives one equation. Step 3 — The final equation comes from equating the ratio of leading coefficients to $L$. In this problem: existence at $x=0$ gives $c = -1$; matching order gives $a = -2$; ratio condition gives $b^2 = 2$. This structured approach guarantees you won’t miss any condition. JEE often awards partial credit for finding even one or two of the constants correctly, so this systematic method is exam-safe.
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FAQs
Q
Why use Taylor series instead of L’Hôpital?
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Taylor series gives all coefficients at once in one expansion, whereas L’Hôpital requires multiple derivatives. For parametric limits with unknowns, Taylor series is faster and gives the conditions on constants directly.
Q
How is c = −1 found?
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At x=0, numerator = 1 + 2 + (c−2) = c+1. Denominator = 0. For finite limit, numerator must also = 0, so c+1 = 0, giving c = −1.
Q
How is a = −2 found?
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Denominator starts at x². So numerator’s x-coefficient must = 0. Coefficient of x in numerator = (a−1)+3 = a+2. Setting a+2 = 0 gives a = −2.
Q
What is the expansion of 2cos(bx)?
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2cos(bx) = 2(1 − b²x²/2 + …) = 2 − b²x² + higher order terms. No x¹ term (cosine is even function).
Q
Why does denominator start at x²?
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x cos x − log(1+x) = (x − x³/2 + …) − (x − x²/2 + x³/3 − …) = x²/2 + …. The x terms cancel, leaving x²/2 as the leading term.
Q
How is b² = 2 derived?
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With a=−2, c=−1: numerator = (3−b²)x² + …, denominator = x²/2 + … Limit = 2(3−b²) = 2, so b² = 2.
Q
What is the x² coefficient in numerator?
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With a = −2: (a−1)²/2 − b² − 3/2 = (−3)²/2 − b² − 3/2 = 9/2 − b² − 3/2 = 3 − b².
Q
Can L’Hôpital be applied here?
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Yes, but you’d need to apply L’Hôpital twice (since both numerator and denominator vanish at x=0 after finding c and a). Taylor series is much more efficient for this type.
Q
What are the final values of a, b, c?
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a = −2, b = ±√2 (b² = 2), c = −1. Note: a² + b² + c² only needs b², so b’s sign doesn’t matter.
Q
What is the correct answer?
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Option D: a² + b² + c² = 4 + 2 + 1 = 7.
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