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Let A = lim(x -> -2) [tan(pi.x)/x+2] + lim (x → inf) (1 + 1/x 2 ) x Prove that A > 4

Let A = lim(x -> -2) [tan(pi.x)/x+2] + lim (x → inf) (1 + 1/x2)x
Prove that A > 4

Grade:11

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:Hello student, please find answer to your question
A = \lim_{x\rightarrow -2}\frac{tan\pi x}{x+2} + \lim_{x\rightarrow \infty }(1 + \frac{1}{x^{2}})^{x}
A_{1} = \lim_{x\rightarrow -2}\frac{tan\pi x}{x+2}
A_{2} = \lim_{x\rightarrow \infty }(1 + \frac{1}{x^{2}})^{x}
A_{1} = \lim_{x\rightarrow -2}\frac{tan\pi x}{x+2}
It is zero by zero form, apply L’Hospital Rule
A_{1} = \lim_{x\rightarrow -2}\frac{(sec^{2}\pi x)\pi }{1}
A_{1} = \frac{(sec^{2}(-2\pi) )\pi }{1} = \pi
A_{2} = \lim_{x\rightarrow \infty }(1 + \frac{1}{x^{2}})^{x}
A_{2} = \lim_{x\rightarrow \infty }(1 + \frac{1}{x^{2}})^{x^{2}.\frac{1}{x}}
\lim_{x\rightarrow \infty }(1 + \frac{1}{x^{2}})^{x^{2}} = e
A_{2} = e^{\lim_{x\rightarrow \infty }\frac{1}{x}}
A_{2} = e^{0} = 1
A = A_{1} + A_{2}
A = \pi + 1 > 4
Hence, Proved

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