Question

prove that.

lim Vx (Vx+ 1 - Vx) = 1/2 X>00 ​

Answer 1

Answer:

The idea is to transform the expression by multiplying $(\sqrt{x + 1} - \sqrt{x})$ with its conjugate, $(\sqrt{x + 1} + \sqrt{x})$.

Step-by-step explanation:

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

The factor $(\sqrt{x + 1} - \sqrt{x})$ is irrational. However, when multiplied with its square root conjugate $(\sqrt{x + 1} + \sqrt{x})$, the product would become rational:

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

The idea is to multiply $\sqrt{x}\, (\sqrt{x + 1} - \sqrt{x})$ by $\displaystyle \frac{\sqrt{x + 1} + \sqrt{x}}{\sqrt{x + 1} + \sqrt{x}}$ so as to make it easier to take the limit.

Since $\displaystyle \frac{\sqrt{x + 1} + \sqrt{x}}{\sqrt{x + 1} + \sqrt{x}} = 1$, multiplying the expression by this fraction would not change the value of the original expression.

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

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 $\sqrt{x}$) would ensure that all but the constant terms would approach $0$ under this limit:

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

By continuity:

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

Answer 2

Answer:

Hello,

Step-by-step explanation:

$\displaystyle \lim_{x \to \infty} \sqrt{x}*(\sqrt{x+1}-\sqrt{x} ) \\\\\\= \lim_{x \to \infty}\dfrac{ \sqrt{x}*(\sqrt{x+1}-\sqrt{x} )*(\sqrt{x+1}+\sqrt{x} )}{\sqrt{x+1} +\sqrt{x} } \\\\= \lim_{x \to \infty} \dfrac{\sqrt{x} *1}{\sqrt{x+1} +\sqrt{x} } \\\\\\= \lim_{x \to \infty} \dfrac{1} {\sqrt {\dfrac {x+1} {x} }+\sqrt{\dfrac{x}{x} } } \\\\\\=\dfrac{1} {\sqrt {1}+\sqrt{1} } \\\\\\=\dfrac{1} {2}$