Solveeit Logo
Login

Question

Mathematics Question on Limits

Let
β=lim⁡x→0αx−(e3x−1)αx(e3x−1)β = \lim_{x →0} \frac{αx-(e^{3x}-1)}{αx(e^{3x}-1) }for someα∈R. α \in R.
Then the value of α+β is

A

145\frac{14}{5}

B

32\frac{3}{2}

C

52\frac{5}{2}

D

72\frac{7}{2}

Answer

52\frac{5}{2}

Explanation

Solution

The correct answer is (C) : 52\frac{5}{2}
β=lim⁡x→0αx−(e3x−1)αx(e3x−1),α∈R.β = \lim _{x →0} \frac{αx-(e^{3x}-1)}{αx(e^{3x}-1)}, α \in R.
=lim⁡x→0α3−(e3x−13x)αx(e3x−13x)=\lim_{x →0} \frac{\frac{α}{3}-(\frac{e^{3x}-1}{3x})}{αx(\frac{e^{3x-1}}{3x})}
So, α = 3 (to make indeterminant form)
β=lim⁡x→01−(e3x−13x)3xβ = \lim_{x\rightarrow0}\frac{1-(\frac{e^{3x}-1}{3x})}{3x}
=1−(3x+9x22+...)3x3x= \frac{1-\frac{(3x+\frac{9x^2}{2}+...)}{3x}}{3x}
=−(92x2+(3x)331+...)9x2=−12= -\frac{(\frac{9}{2}x^2+\frac{(3x)^3}{31}+...)}{9x^2}=\frac{-1}{2}
∴α+β=3−12∴ α+β = 3 -\frac{1}{2}
=52= \frac{5}{2}