- lim x - - x 2 - x - 1 - - x 2 - 1 solve - -

lim_x→∞√(x^2+x+1) √(x^2+1)×(√(x^2+x+1)+√(x^2+1)/√(x^2+x+1)+√(x^2+1)) =lim_x→∞((x^2+x+1) (x^2+1)/√(x^2+x+1)+√(x^2+1)) =lim_x→∞(x/√(x^2+x+1)+√(x^2+1) ...

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Byju's Answer

Standard XII

Mathematics

Algebra of Limits

= lim x →∞ √...

Question

Solve: ${lim}_{x \to \infty} (\sqrt{x^{2} + x + 1} - \sqrt{x^{2} + 1})$

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Solution

${lim}_{x \to \infty} (\sqrt{x^{2} + x + 1} - \sqrt{x^{2} + 1}) \times (\frac{\sqrt{x^{2} + x + 1} + \sqrt{x^{2} + 1}}{\sqrt{x^{2} + x + 1} + \sqrt{x^{2} + 1}}) = {lim}_{x \to \infty} (\frac{(x^{2} + x + 1) - (x^{2} + 1)}{\sqrt{x^{2} + x + 1} + \sqrt{x^{2} + 1}}) [∵ a^{2} - b^{2} = (a + b) (a - b)] = {lim}_{x \to \infty} (\frac{x}{\sqrt{x^{2} + x + 1} + \sqrt{x^{2} + 1}}) = {lim}_{x \to \infty} ⎛ ⎜ ⎜ ⎜ ⎝ \frac{\frac{x}{x}}{\frac{\sqrt{x^{2} + x + 1} + \sqrt{x^{2} + 1}}{x}} ⎞ ⎟ ⎟ ⎟ ⎠ = {lim}_{x \to \infty} ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{1}{\sqrt{1 + (\frac{1}{x}) + (\frac{1}{x^{2}})} + \sqrt{1 + (\frac{1}{x^{2}})}} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ = (\frac{1}{1 + 1}) = \frac{1}{2}$

Suggest Corrections

Solve: limx→∞(√x2+x+1−√x2+1)

Solve: ${lim}_{x \to \infty} (\sqrt{x^{2} + x + 1} - \sqrt{x^{2} + 1})$