Show that limit of $\sqrt{x} {\sqrt{x + 2} - \sqrt{x}} = 1$ $sqrt(x){sqrt(x+2)-sqrt(x)}=1$ as x approaches to infinity?

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Answer 1 · 2017-12-08T14:25:06

$lim_(xrarroo) sqrt(x)(sqrt(x+2)-sqrt(x))$ has indeterminate form $oo(oo-oo)$

Use a conjugate to rewrite the expression.

$(sqrt(x)(sqrt(x+2)-sqrt(x)))/1 * ((sqrt(x+2)+sqrt(x)))/((sqrt(x+2)+sqrt(x))) = (sqrtx((x+2)-x))/(sqrt(x+2)+sqrt(x))$

$= (2sqrtx)/(sqrt(x+2)+sqrt(x))$

Now the limit has indeterminate form $oo/oo$ . Remove a common factor of $sqrtx$

$= (cancel(sqrtx)(2))/(cancel(sqrtx)(sqrt(1+2/sqrtx)+1))$

The limit as $xrarroo$ is $2/(sqrt(1+0)+1) = 2/2 = 1$

Show that limit of sqrt(x){sqrt(x+2)-sqrt(x)}=1√x{√x+2−√x}=1sqrt(x){sqrt(x+2)-sqrt(x)}=1 as x approaches to infinity?