Lim x -x2-8 x-3-x2-4 x-3-

We have, lim_x→∞(√((x^2+8x+3)) √((x^2+4x+3))) If we directly substitute x=∞ , we find that it is an ∞ ∞ form. Rationalize it by (√((x^2+8x+3))+ √((x^2+4x+3))) ...

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Byju's Answer

Standard XI

Mathematics

Factorization Method Form to Remove Indeterminate Form

lim x →∞√x2+8...

Question

$lim x \to \infty (\sqrt{(x^{2} + 8 x + 3)} - \sqrt{(x^{2} + 4 x + 3)}) =$

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Solution

we have,

$lim x \to \infty (\sqrt{(x^{2} + 8 x + 3)} - \sqrt{(x^{2} + 4 x + 3)})$

If we directly substitute $x = \infty$ , we find that it is an $\infty - \infty$ form.

Rationalize it by $(\sqrt{(x^{2} + 8 x + 3)} + \sqrt{(x^{2} + 4 x + 3)})$

$lim x \to \infty \frac{(\sqrt{(x^{2} + 8 x + 3)} - \sqrt{(x^{2} + 4 x + 3)}) (\sqrt{(x^{2} + 8 x + 3)} + \sqrt{(x^{2} + 4 x + 3)})}{(\sqrt{(x^{2} + 8 x + 3)} + \sqrt{(x^{2} + 4 x + 3)})}$

= $lim x \to \infty \frac{x^{2} + 8 x - 3 - (x^{2} + 4 x + 3)}{\sqrt{(x^{2} + 8 x + 3)} + \sqrt{(x^{2} + 4 x + 3)}}$

= $lim x \to \infty \frac{4 x}{\sqrt{(x^{2} + 8 x + 3)} + \sqrt{(x^{2} + 4 x + 3)}}$

Degree of polynomial in numerator and denominator is same as 1.

dividing numerator and denominator by

$lim x \to \infty \frac{4}{(\sqrt{1 + \frac{8}{x} + \frac{3}{x^{2}}} + \sqrt{1 + \frac{4}{x} + \frac{3}{x^{2}}})}$

= $\frac{4}{\sqrt{1 + 0 + 0} + \sqrt{1 + 0 + 0}}$

= $\frac{4}{1 + 1} = \frac{4}{2} = 2$

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limx→∞(√(x2+8x+3)−√(x2+4x+3))=

$lim x \to \infty (\sqrt{(x^{2} + 8 x + 3)} - \sqrt{(x^{2} + 4 x + 3)}) =$