The value of `lim_(x->0)((sinx-tanx)^2-(1-cos2x)^4+x^5)/(7(tan^(- 1)x)^7+(sin^(- 1)x)^6+3sin^5x)` equal to :

Options

A. 0
B. 1
C. 2
D. `1/3`

Answer

To solve the limit problem

\[ \lim_{x \to 0} \frac{(\sin x - \tan x)^2 - (1 - \cos 2x)^4 + x^5}{7(\tan^{-1} x)^7 + (\sin^{-1} x)^6 + 3\sin^5 x} \]

we will follow these steps:

### Step 1: Simplify the Numerator
We start with the numerator:

\[ (\sin x - \tan x)^2 - (1 - \cos 2x)^4 + x^5 \]

Using the identity \(\tan x = \frac{\sin x}{\cos x}\), we can rewrite \(\sin x - \tan x\):

\[ \sin x - \tan x = \sin x - \frac{\sin x}{\cos x} = \sin x \left(1 - \frac{1}{\cos x}\right) = \sin x \left(\frac{\cos x - 1}{\cos x}\right) \]

Using the small angle approximation, we know that as \(x \to 0\), \(\sin x \approx x\) and \(\cos x \approx 1 - \frac{x^2}{2}\). Thus,

\[ \cos x - 1 \approx -\frac{x^2}{2} \]

So,

\[ \sin x - \tan x \approx \frac{x \cdot (-\frac{x^2}{2})}{1} = -\frac{x^3}{2} \]

Now squaring this gives:

\[ (\sin x - \tan x)^2 \approx \left(-\frac{x^3}{2}\right)^2 = \frac{x^6}{4} \]

Next, we simplify \(1 - \cos 2x\):

\[ 1 - \cos 2x = 2\sin^2 x \approx 2\left(\frac{x^2}{2}\right) = x^2 \]

Thus,

\[ (1 - \cos 2x)^4 \approx (x^2)^4 = x^8 \]

Now substituting back into the numerator:

\[ \frac{x^6}{4} - x^8 + x^5 = \frac{x^6}{4} + x^5 - x^8 \]

### Step 2: Simplify the Denominator
Now, we simplify the denominator:

\[ 7(\tan^{-1} x)^7 + (\sin^{-1} x)^6 + 3\sin^5 x \]

Using the small angle approximations:

\[ \tan^{-1} x \approx x \quad \text{and} \quad \sin^{-1} x \approx x \]

Thus,

\[ 7(\tan^{-1} x)^7 \approx 7x^7, \quad (\sin^{-1} x)^6 \approx x^6, \quad \text{and} \quad 3\sin^5 x \approx 3x^5 \]

So the denominator becomes:

\[ 7x^7 + x^6 + 3x^5 \]

### Step 3: Combine and Factor
Now we can rewrite our limit:

\[ \lim_{x \to 0} \frac{\frac{x^6}{4} + x^5 - x^8}{7x^7 + x^6 + 3x^5} \]

Factoring out \(x^5\) from both the numerator and denominator:

Numerator:

\[ x^5\left(\frac{x}{4} + 1 - x^3\right) \]

Denominator:

\[ x^5(7x^2 + x + 3) \]

Thus, the limit simplifies to:

\[ \lim_{x \to 0} \frac{\frac{x}{4} + 1 - x^3}{7x^2 + x + 3} \]

### Step 4: Evaluate the Limit
Now we can substitute \(x = 0\):

Numerator:

\[ \frac{0}{4} + 1 - 0 = 1 \]

Denominator:

\[ 7(0)^2 + 0 + 3 = 3 \]

Thus, the limit evaluates to:

\[ \frac{1}{3} \]

### Final Answer
The value of

\[ \lim_{x \to 0} \frac{(\sin x - \tan x)^2 - (1 - \cos 2x)^4 + x^5}{7(\tan^{-1} x)^7 + (\sin^{-1} x)^6 + 3\sin^5 x} = \frac{1}{3} \]