with its conjugate, 
.
and 
, 
.
.
by 
so as to make it easier to take the limit. 
, multiplying the expression by this fraction would not change the value of the original expression.![\begin{aligned} & \lim\limits_{x \to \infty} \sqrt{x} \, (\sqrt{x + 1} - \sqrt{x}) \\ &= \lim\limits_{x \to \infty} \left[\sqrt{x} \, (\sqrt{x + 1} - \sqrt{x})\cdot \frac{\sqrt{x + 1} + \sqrt{x}}{\sqrt{x + 1} + \sqrt{x}}\right] \\ &= \lim\limits_{x \to \infty} \frac{\sqrt{x}\, ((x + 1) - x)}{\sqrt{x + 1} + \sqrt{x}} \\ &= \lim\limits_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x + 1}+ \sqrt{x}}\end{aligned}](/tpl/images/1396/1153/477e5.png)
.
in both the numerator and the denominator are now both 
. Hence, dividing both the numerator and the denominator by 
(same as 
) would ensure that all but the constant terms would approach 
under this limit:
.
.
Prove that.
lim Vx (Vx+ 1 - Vx) = 1/2 X>00
...
Mathematics, 19.07.2021 15:20 elopezhilario4389
Prove that.
lim Vx (Vx+ 1 - Vx) = 1/2 X>00
Answers: 1
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