JEE Mains · Maths · STD 11 - 12. limits
Let \(\beta=\lim _{x \rightarrow 0} \frac{\alpha x-\left(e^{3 x}-1\right)}{\alpha x\left(e^{3 x}-1\right)}\) for some \(\alpha \in R\). Then the value of \(\alpha+\beta\) is.
- A \(\frac{14}{5}\)
- B \(\frac{3}{2}\)
- C \(\frac{5}{2}\)
- D \(\frac{7}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{5}{2}\)
Step-by-step Solution
Detailed explanation
\(\beta=\lim _{x \rightarrow 0} \frac{\alpha x-\left(e^{3 x}-1\right)}{\alpha x\left(e^{3 x}-1\right)}\) \(\beta=\lim _{x \rightarrow 0} \frac{1+\alpha x-\left[1+3 x+\frac{9 x^{2}}{2 !}+\ldots .\right]}{(\alpha x) \frac{\left(e^{3 x}-1\right)}{3 x} 3 x}\)…
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