Question

lim x tends to infinity (√x²+x-1 -x)

Answer 1

Answer:

1/2

Step-by-step explanation:

Hope I can clear your doubt

Answer 2

As limit x tends to ∞, ($\sqrt{x^{2} +x-1}$  - x) tends to 1/2.

Given:

($\sqrt{x^{2} +x-1}$  - x)

To Find:

$\lim_{x \to \infty}$  ($\sqrt{x^{2} +x-1}$  - x) =?

Solution:

Rationalizing ($\sqrt{x^{2} +x-1}$  - x) , we get

($\sqrt{x^{2} +x-1}$  - x) =  × $\frac{\sqrt{x^{2} +x-1}+x}{\sqrt{x^{2} +x-1}+ x}$

⇒ ($\sqrt{x^{2} +x-1}$  - x) =$\frac{({x^{2} +x-1})^{2} -x^{2} }{\sqrt{x^{2} +x-1}+ x}$

⇒ ($\sqrt{x^{2} +x-1}$  - x) = $\frac{{x} -1 }{\sqrt{x^{2} +x-1}+ x}$

Dividing numerator and denominator by x, we have

($\sqrt{x^{2} +x-1}$  - x) = $\frac{1-\frac{1}{x} } {\sqrt{1 +\frac{1}{x} -\frac{1}{x^{2}} }+ 1}$        ......................(I)

We know that as x takes on larger values, 1/x tends to 0, i.e.

$\lim_{x \to \infty} \frac{1}{x}$ = 0

$\lim_{x \to \infty} \frac{1}{x^{2} }$ = 0

Hence, as x tends to ∞, equation (I) becomes

$\lim_{x \to \infty}$  ($\sqrt{x^{2} +x-1}$  - x) = $\lim_{x \to \infty}$  $\frac{1-\frac{1}{x} } {\sqrt{1 +\frac{1}{x} -\frac{1}{x^{2}} }+ 1}$ = (1 - 0)/ (√1 + 1) = 1/2 .

∴ $\lim_{x \to \infty}$  ($\sqrt{x^{2} +x-1}$  - x) = 1/2

Hence, As limit x tends to ∞, ($\sqrt{x^{2} +x-1}$  - x) tends to 1/2.

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