Question

Prove that.

lim Vx (Vx+ 1 - Vx) = 1/2 X>00 ​

Answer 1

Hello,

Explanation:

`\displaystyle \lim_(x \to \infty) √(x)*(√(x+1)-√(x) ) \\\\\\= \lim_(x \to \infty)( √(x)*(√(x+1)-√(x) )*(√(x+1)+√(x) ))/(√(x+1) +√(x) ) \\\\= \lim_(x \to \infty) (√(x) *1)/(√(x+1) +√(x) ) \\\\\\= \lim_(x \to \infty) \frac{1} {\sqrt {\frac {x+1} {x} }+\sqrt{(x)/(x) } } \\\\\\=\frac{1} {\sqrt {1}+√(1) } \\\\\\=\frac{1} {2} \\`

Answer 2

The idea is to transform the expression by multiplying
`(√(x + 1) - √(x))` with its conjugate,
`(√(x + 1) + √(x))`.

Explanation:

For any real number
`a` and
`b`,
`(a + b)\, (a - b) = a^(2) - b^(2)`.

The factor
`(√(x + 1) - √(x))` is irrational. However, when multiplied with its square root conjugate
`(√(x + 1) + √(x))`, the product would become rational:

`\begin{aligned} & (√(x + 1) - √(x)) \, (√(x + 1) + √(x)) \\ &= (√(x + 1))^(2) -(√(x))^(2) \\ &= (x + 1) - (x) = 1\end{aligned}`.

The idea is to multiply
`√(x)\, (√(x + 1) - √(x))` by
`\displaystyle (√(x + 1) + √(x))/(√(x + 1) + √(x))` so as to make it easier to take the limit.

Since
`\displaystyle (√(x + 1) + √(x))/(√(x + 1) + √(x)) = 1`, multiplying the expression by this fraction would not change the value of the original expression.

`\begin{aligned} & \lim\limits_(x \to \infty) √(x) \, (√(x + 1) - √(x)) \\ &= \lim\limits_(x \to \infty) \left[√(x) \, (√(x + 1) - √(x))\cdot (√(x + 1) + √(x))/(√(x + 1) + √(x))\right] \\ &= \lim\limits_(x \to \infty) (√(x)\, ((x + 1) - x))/(√(x + 1) + √(x)) \\ &= \lim\limits_(x \to \infty) (√(x))/(√(x + 1)+ √(x))\end{aligned}`.

The order of
`x` in both the numerator and the denominator are now both
`(1/2)`. Hence, dividing both the numerator and the denominator by
`x^((1/2))` (same as
`√(x)`) would ensure that all but the constant terms would approach
`0` under this limit:

`\begin{aligned} & \lim\limits_(x \to \infty) √(x) \, (√(x + 1) - √(x)) \\ &= \cdots\\ &= \lim\limits_(x \to \infty) (√(x))/(√(x + 1)+ √(x)) \\ &= \lim\limits_(x \to \infty) (√(x) / √(x))/((√(x + 1) / √(x)) + (√(x) / √(x))) \\ &= \lim\limits_(x \to \infty)(1)/(√((x / x) + (1 / x)) + 1) \\ &= \lim\limits_(x \to \infty) (1)/(√(1 + (1/x)) + 1)\end{aligned}`.

By continuity:

`\begin{aligned} & \lim\limits_(x \to \infty) √(x) \, (√(x + 1) - √(x)) \\ &= \cdots\\ &= \lim\limits_(x \to \infty) (√(x))/(√(x + 1)+ √(x)) \\ &= \cdots \\ &= \lim\limits_(x \to \infty) (1)/(√(1 + (1/x)) + 1) \\ &= \frac{1}{\sqrt{1 + \lim\limits_(x \to \infty)(1/x)} + 1} \\ &= (1)/(1 + 1) \\ &= (1)/(2)\end{aligned}`.