Let
\(β = \lim_{x →0} \frac{αx-(e^{3x}-1)}{αx(e^{3x}-1) }\)for some\( α \in R.\)
Then the value of α+β is
Let
\(β = \lim_{x →0} \frac{αx-(e^{3x}-1)}{αx(e^{3x}-1) }\)for some\( α \in R.\)
Then the value of α+β is
\(\frac{14}{5}\)
\(\frac{3}{2}\)
\(\frac{5}{2}\)
\(\frac{7}{2}\)
The Correct Option is C
Solution and Explanation
The correct answer is (C) : \(\frac{5}{2}\)
\(β = \lim _{x →0} \frac{αx-(e^{3x}-1)}{αx(e^{3x}-1)}, α \in R.\)
\(=\lim_{x →0} \frac{\frac{α}{3}-(\frac{e^{3x}-1}{3x})}{αx(\frac{e^{3x-1}}{3x})}\)
So, α = 3 (to make indeterminant form)
\(β = \lim_{x\rightarrow0}\frac{1-(\frac{e^{3x}-1}{3x})}{3x}\)
\(= \frac{1-\frac{(3x+\frac{9x^2}{2}+...)}{3x}}{3x}\)
\(= -\frac{(\frac{9}{2}x^2+\frac{(3x)^3}{31}+...)}{9x^2}=\frac{-1}{2}\)
\(∴ α+β = 3 -\frac{1}{2}\)
\(= \frac{5}{2}\)
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Concepts Used:
Limits
A function's limit is a number that a function reaches when its independent variable comes to a certain value. The value (say a) to which the function f(x) approaches casually as the independent variable x approaches casually a given value "A" denoted as f(x) = A.
If limx→a- f(x) is the expected value of f when x = a, given the values of ‘f’ near x to the left of ‘a’. This value is also called the left-hand limit of ‘f’ at a.
If limx→a+ f(x) is the expected value of f when x = a, given the values of ‘f’ near x to the right of ‘a’. This value is also called the right-hand limit of f(x) at a.
If the right-hand and left-hand limits concur, then it is referred to as a common value as the limit of f(x) at x = a and denote it by lim x→a f(x).