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

Ans:Hello student, please find answer to your questionIt is zero by zero form, apply L’Hospital Rule 4 src=https://files.askiitians.com/cdn1/cms-content/common/latex.codecogs.comgif.latexa_pi1gt4.jpg>Hence, Proved

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Differential Calculus> Let A = lim(x -> -2) [tan(pi.x)/x+2] + li...

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

Kawal , 10 Years ago

Grade 11

1 Answers

Jitender Singh

Last Activity: 10 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