Horrible Limit

Calculus Level 5

lim ⁡ x → ∞ ( x 2 + 2 x + 3 − x 2 + 3 ) x \large \lim_{x \to \infty} \left( \sqrt{x^2+2x+3} - \sqrt{x^2+3} \right)^x

Calculate the above limit to 3 decimal places.


The answer is 0.606.

Vilakshan Gupta by Vilakshan Gupta , 19, India

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2 solutions

Chew-Seong Cheong
Mar 31, 2018

L = lim ⁡ x → ∞ ( x 2 + 2 x + 3 − x 2 + 3 ) x = lim ⁡ x → ∞ ( x 1 + 2 x + 3 x 2 − x 1 + 3 x 2 ) x = lim ⁡ x → ∞ x x ( ( 1 + 2 x + 3 x 2 ) 1 2 − ( 1 + 3 x 2 ) 1 2 ) x By binomial expansion = lim ⁡ x → ∞ x x ( ( 1 + 1 2 ( 2 x + 3 x 2 ) − 1 8 ( 2 x + 3 x 2 ) 2 + ⋯ ) − ( 1 + 1 2 ( 3 x 2 ) − 1 8 ( 3 x 2 ) 2 + ⋯ ) ) x = lim ⁡ x → ∞ x x ( 1 x − 1 2 x 2 + O ( x − 3 ) ) x = lim ⁡ x → ∞ ( 1 − 1 2 x + O ( x − 2 ) ) − 2 x ( − 1 2 ) = e − 1 2 = 1 e ≈ 0.606 \begin{aligned} L & = \lim_{x \to \infty} \left(\sqrt{x^2+2x+3} - \sqrt{x^2+3}\right)^x \\ & = \lim_{x \to \infty} \left(x\sqrt{1+\frac 2x+\frac 3{x^2}} - x\sqrt{1+\frac 3{x^2}}\right)^x \\ & = \lim_{x \to \infty} x^x\left({\color{#3D99F6}\left(1+\frac 2x+\frac 3{x^2}\right)^\frac 12 - \left(1+\frac 3{x^2}\right)^\frac 12}\right)^x & \small \color{#3D99F6} \text{By binomial expansion} \\ & = \lim_{x \to \infty} x^x\left({\color{#3D99F6}\left(1+\frac 12 \left(\frac 2x + \frac 3{x^2}\right)-\frac 18\left(\frac 2x + \frac 3{x^2}\right)^2+\cdots \right)-\left(1+\frac 12 \left(\frac 3{x^2}\right)-\frac 18\left(\frac 3{x^2}\right)^2+\cdots \right)}\right)^x \\ & = \lim_{x \to \infty} x^x\left(\frac 1x - \frac 1{2x^2} + O(x^{-3}) \right)^x \\ & = \lim_{x \to \infty} \left(1 - \frac 1{2x} + O(x^{-2}) \right)^{-2x \left(-\frac 12 \right)} \\ & = e^{-\frac 12} = \frac 1{\sqrt e} \approx \boxed{0.606} \end{aligned}

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Naren Bhandari
Mar 27, 2018

L = lim ⁡ x → ∞ ( x 2 + 2 x + 3 − x 2 + 3 ) x L = lim ⁡ x → ∞ ( 2 x x 2 + 2 x + 3 + x 2 + 3 ) x L = lim ⁡ x → ∞ ( x 2 + 2 x + 3 + x 3 + 3 + 2 x − ( x 2 + 2 x + 3 + x 2 + 3 ) x 2 + 2 x + 3 + x 2 + 3 ) x L = lim ⁡ x → ∞ ( 1 + 2 x − ( x 2 + 2 x + 3 + x 2 + 3 ) x 2 + 2 x + 3 + x 2 + 3 ) x L = lim ⁡ x → ∞ ( 1 + 2 x − x 2 + 2 x + 3 − x 2 + 3 x 2 + 2 x + 3 + x 2 + 3 ) x L = lim ⁡ x → ∞ ( 1 + 1 x 2 + 2 x + 3 + x 2 + 3 2 x − x 2 + 2 x + 3 − x 2 + 3 ) x x 2 + 2 x + 3 + x 2 + 3 2 x − x 2 − 2 x + 3 − x 2 + 3 . 2 x − x 2 + 2 x + 3 − x 2 + 3 x 2 + 2 x + 3 + x 2 + 3 L = exp ( lim ⁡ n → ∞ x ( 2 x − x 2 + 2 x + 3 − x 2 + 3 ) x 2 + 2 x + 3 + x 2 + 3 ) L = exp ( − 1 2 ) = 1 e \begin{aligned} & L = \displaystyle\lim_{x\to \infty}\left(\sqrt{x^2 +2x+3} -\sqrt{x^2 +3} \right)^x\\& L = \displaystyle\lim_{x\to \infty}\left(\dfrac{2x}{\sqrt{x^2+2x+3} + \sqrt{x^2+3}}\right)^x\\& L = \displaystyle\lim_{x\to \infty}\left(\dfrac{\sqrt{x^2+2x+3}+\sqrt{x^3+3}+ 2x -(\sqrt{x^2+2x+3}+\sqrt{x^2+3})}{\sqrt{x^2+2x+3}+ \sqrt{x^2+3}}\right)^x \\& L = \displaystyle \lim_{x\to\infty} \left(1 + \dfrac{2x-(\sqrt{x^2+2x+3} +\sqrt{x^2+3})}{\sqrt{x^2+2x+3}+\sqrt{x^2+3}}\right)^x \\& L = \displaystyle\lim_{x\to\infty} \left(1 + \dfrac{2x -\sqrt{x^2+2x+3} -\sqrt{x^2+3}}{\sqrt{x^2+2x+3}+\sqrt{x^2+3}}\right)^x\\& L = \displaystyle\lim_{x\to\infty}\left(1+\dfrac{1}{\dfrac{\sqrt{x^2+2x+3}+\sqrt{x^2+3}}{2x-\sqrt{x^2+2x+3}-\sqrt{x^2+3}}}\right)^{\small\dfrac{x\sqrt{x^2+2x+3}+\sqrt{x^2+3}}{2x-\sqrt{x^2-2x+3}-\sqrt{x^2+3}}.\dfrac{2x-\sqrt{x^2+2x+3}-\sqrt{x^2+3}}{\sqrt{x^2+2x+3}+\sqrt{x^2+3}}}\\& L = \text{exp}\left(\displaystyle\lim_{n\to\infty}\dfrac{x(2x -\sqrt{x^2+2x+3}-\sqrt{x^2+3})}{\sqrt{x^2+2x+3}+\sqrt{x^2+3}}\right)\\& L = \text{exp}\left(-\dfrac{1}{2}\right) = \dfrac{1}{\sqrt e} \end{aligned}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

I have got it.

Chew-Seong Cheong - 3 years, 2 months ago

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Wow !! Upvoted :) Sir . I hope that you liked my solution as well. :D

Naren Bhandari - 3 years, 2 months ago

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