If x- 0lim- xex- loge1-x- x2e-xx sin2x-10- - - - - -then the value of - - - is

Using expansion, we have: lim_x→ 0 α x ( 1+x1! +x^22!+⋯ ) β ( x x^22 +x^33+⋯ )+γ x^2 ( 1 x1!+x^22! +... )x^3 ( sin xx )^2=10 ⇒lim_x→ 0 nbsp; x(α β )+x^2 ( α ...

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Byju's Answer

Standard XII

Mathematics

Integration to Solve Modified Sum of Binomial Coefficients

If x→ 0limα x...

Question

If $lim x \to 0 \frac{α x e^{x} - β {log}_{e} (1 + x) + γ x^{2} e^{- x}}{x {sin}^{2} x} = 10, α, β, γ \in R,$
then the value of $α + β + γ$ is

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Solution

Using expansion, we have:
$lim x \to 0 \frac{α x (1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \dots) - β (x - \frac{x^{2}}{2} + \frac{x^{3}}{3} + \dots) + γ x^{2} (1 - \frac{x}{1!} + \frac{x^{2}}{2!} + . . .)}{x^{3} {(\frac{sin x}{x})}^{2}} = 10$

$\Rightarrow lim x \to 0 \frac{x (α - β) + x^{2} (α + \frac{β}{2} + γ) + x^{3} (\frac{α}{2} - \frac{β}{3} - γ) + . . .}{x^{3}} = 10$

$\Rightarrow α - β = 0, α + \frac{β}{2} + γ = 0, \frac{α}{2} - \frac{β}{3} - γ = 10$

$\Rightarrow α = 6, β = 6, γ = - 9$

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If limx→0αxex−β loge(1+x)+γx2e−xxsin2x=10,α,β,γ∈R, then the value of α+β+γ is

Using expansion, we have: limx→0 αx(1+x1!+x22!+⋯)−β(x−x22+x33+⋯)+γx2(1−x1!+x22!+...)x3(sinxx)2=10 ⇒limx→0x(α−β)+x2(α+β2+γ)+x3(α2−β3−γ)+...x3=10 ⇒α−β=0,α+β2+γ=0,α2−β3−γ=10 ⇒α=6,β=6,γ=−9

If $lim x \to 0 \frac{α x e^{x} - β {log}_{e} (1 + x) + γ x^{2} e^{- x}}{x {sin}^{2} x} = 10, α, β, γ \in R,$
then the value of $α + β + γ$ is