Question

$evaluate \: \: lim \: x - - > \infty \: \: 1 - \sqrt{x } \div 1 + \sqrt{x}$
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Answer 1

$\displaystyle\text{This limit can be calculated using the result,}\:\:\: \boxed{\lim_{x \to \infty} \frac{1}{\sqrt{x}}=0}\\\\\\\lim_{x \to \infty} \frac{1-\sqrt{x}}{1+\sqrt{x}}\\\\\\= \lim_{x \to \infty} \frac{\sqrt{x}\left(1-\dfrac{1}{\sqrt{x}\right)}}{\sqrt{x}\left(1+\dfrac{1}{\sqrt{x}}\right)}\\\\\\=\dfrac{\displaystyle\lim_{x \to \infty}(1)-\lim_{x \to \infty}\left( \dfrac{1}{\sqrt{x}\right)}}{\displaystyle \lim_{x \to \infty} (1)+ \lim_{x \to \infty} \left(\dfrac{1}{\sqrt{x}\right)}}=\dfrac{1-0}{1-0}=1$