MHT CET · Maths · Limits
\(\lim _{x \rightarrow \infty}\left[\frac{8 x^2+5 x+3}{2 x^2-7 x-5}\right]^{\frac{4 x+3}{8 x-1}}=\)
- A 4
- B \(\frac{1}{2}\)
- C 2
- D \(\sqrt{2}\)
Answer & Solution
Correct Answer
(C) 2
Step-by-step Solution
Detailed explanation
\(\lim _{x \rightarrow \infty}\left(\frac{8 x^2+5 x+3}{2 x^2-7 x-5}\right)^{\frac{4 x+3}{8 x-1}}=\lim _{x \rightarrow \infty}\left(\frac{8+\frac{5}{x}+\frac{3}{x^2}}{2-\frac{7}{x}-\frac{5}{x}}\right)^{\frac{4+\frac{3}{x}}{8-\frac{1}{x}}}\) \(=\left(\frac{8+0+0}{2-0-0}\right)^{\frac{4+0}{8-0}}\)
\(=4^{\frac{1}{2}}=2\)
\(=4^{\frac{1}{2}}=2\)
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