A calculus problem by Sanad Kadu

Calculus Level 4

Find lim ⁡ x → ∞ x − x 2 + x + 1 ln ⁡ ( e x + x ) x \displaystyle \lim_{x\to\infty}x-\sqrt{x^2+x+1}\dfrac {\ln(e^x+x)}{x} .

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Sanad Kadu by Sanad Kadu , 20, India

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1 solution

Sanad Kadu
May 24, 2018

x 2 + x + 1 = x ( 1 + 1 / x + 1 / x 2 ) 1 2 = x + 1 2 + 1 2 x + o ( x − 1 ) \sqrt{x^2+x+1}=x(1+1/x+1/x^2)^\frac12=x+\frac12+\frac1{2x}+o(x^{-1}) ln ⁡ ( e x + x ) x = ln ⁡ e x + ln ⁡ ( 1 + x / e x ) x = 1 + 1 e x + o ( e − x ) \frac{\ln(e^x+x)}x=\frac{\ln e^x+\ln (1+x/e^x)}x=1+\frac1{e^x}+o(e^{-x}) x − x 2 + x + 1 ln ⁡ ( e x + x ) x = x − x − 1 2 − 1 2 x + o ( x − 1 ) → − 1 2 x-\sqrt{x^2+x+1}\frac {\ln(e^x+x)}{x}=x-x-\frac12-\frac1{2x}+o(x^{-1})\to -\frac12

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