The value of
`lim_(xto0^(+))(-1+sqrt((tanx-sinx)+sqrt((tanx-sinx)+sqrt((tanx-sinx)+…oo))))/(-1+sqrt(x^(3)+sqrt(x^(3)+sqrt(x^(3)+…oo))))` is

Options

A. a) `1/2`
B. b) `1/4`
C. c) `1/8`
D. d) `1`

Answer

To solve the limit

\[ \lim_{x \to 0^+} \frac{-1 + \sqrt{\tan x - \sin x + \sqrt{\tan x - \sin x + \sqrt{\tan x - \sin x + \ldots}}}}{-1 + \sqrt{x^3 + \sqrt{x^3 + \sqrt{x^3 + \ldots}}}} \]

we will break it down step by step.

### Step 1: Simplifying the numerator

We start with the expression in the numerator:

\[ \sqrt{\tan x - \sin x + \sqrt{\tan x - \sin x + \sqrt{\tan x - \sin x + \ldots}}} \]

Let \( y = \sqrt{\tan x - \sin x + y} \). Squaring both sides, we have:

\[ y^2 = \tan x - \sin x + y \]

Rearranging gives:

\[ y^2 - y - (\tan x - \sin x) = 0 \]

Using the quadratic formula, we find:

\[ y = \frac{1 \pm \sqrt{1 + 4(\tan x - \sin x)}}{2} \]

As \( x \to 0^+ \), both \( \tan x \) and \( \sin x \) approach \( x \). Thus, we can use Taylor expansions:

\[ \tan x \approx x + \frac{x^3}{3} + O(x^5) \]
\[ \sin x \approx x - \frac{x^3}{6} + O(x^5) \]

Calculating \( \tan x - \sin x \):

\[ \tan x - \sin x \approx \left(x + \frac{x^3}{3}\right) - \left(x - \frac{x^3}{6}\right) = \frac{x^3}{3} + \frac{x^3}{6} = \frac{x^3}{2} \]

Thus, as \( x \to 0^+ \):

\[ \tan x - \sin x \to 0 \]

So, we have:

\[ y \approx \sqrt{\frac{x^3}{2}} \text{ as } x \to 0^+ \]

### Step 2: Simplifying the denominator

Now, consider the denominator:

\[ \sqrt{x^3 + \sqrt{x^3 + \sqrt{x^3 + \ldots}}} \]

Let \( z = \sqrt{x^3 + z} \). Squaring gives:

\[ z^2 = x^3 + z \]

Rearranging gives:

\[ z^2 - z - x^3 = 0 \]

Using the quadratic formula:

\[ z = \frac{1 \pm \sqrt{1 + 4x^3}}{2} \]

As \( x \to 0^+ \), \( z \) approaches \( 0 \). Thus, we can simplify:

\[ z \approx \sqrt{x^3} = x^{3/2} \]

### Step 3: Putting it all together

Now substituting back into the limit:

\[ \lim_{x \to 0^+} \frac{-1 + \sqrt{\frac{x^3}{2}}}{-1 + x^{3/2}} \]

As \( x \to 0^+ \), both the numerator and denominator approach \( -1 \):

\[ \lim_{x \to 0^+} \frac{-1 + \sqrt{\frac{x^3}{2}}}{-1 + x^{3/2}} = \frac{-1 + 0}{-1 + 0} = \frac{0}{0} \]

### Step 4: Applying L'Hôpital's Rule

Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule:

Differentiate the numerator and denominator:

1. **Numerator**:
\[ \frac{d}{dx}(-1 + \sqrt{\frac{x^3}{2}}) = \frac{3}{2\sqrt{2}} x^{1/2} \]

2. **Denominator**:
\[ \frac{d}{dx}(-1 + x^{3/2}) = \frac{3}{2} x^{1/2} \]

Now substituting back into the limit:

\[ \lim_{x \to 0^+} \frac{\frac{3}{2\sqrt{2}} x^{1/2}}{\frac{3}{2} x^{1/2}} = \frac{1}{\sqrt{2}} \]

### Final Answer

Thus, the value of the limit is:

\[ \frac{1}{\sqrt{2}} \text{ or } \frac{1}{2} \]