Examine the continuity of the following functions In case discontinuity identify the type of discontinuity Redifined the function in case of F(x)=(√x²+1 -1)log(1+x)½÷(Tan 2x)³ , x≠0=16 , x=0

To analyze the continuity of the function \( F(x) = \frac{(\sqrt{x^2 + 1} - 1) \log(1 + x)^{1/2}}{(\tan(2x))^3} \) for \( x \neq 0 \) and \( F(0) = 16 \), we need to check the behavior of the function as \( x \) approaches 0 and compare it to the value of the function at \( x = 0 \).

Step 1: Evaluate the Limit as x Approaches 0

First, we need to find the limit of \( F(x) \) as \( x \) approaches 0. This involves evaluating the expression:

\( \lim_{x \to 0} \frac{(\sqrt{x^2 + 1} - 1) \log(1 + x)^{1/2}}{(\tan(2x))^3} \)

Breaking Down the Components

• Numerator: As \( x \) approaches 0, \( \sqrt{x^2 + 1} \) approaches 1. Therefore, \( \sqrt{x^2 + 1} - 1 \) approaches 0. The logarithmic term \( \log(1 + x) \) also approaches 0, which means \( \log(1 + x)^{1/2} \) approaches 0 as well.
• Denominator: The term \( \tan(2x) \) approaches \( \tan(0) = 0 \). Hence, \( (\tan(2x))^3 \) also approaches 0.

Both the numerator and denominator approach 0, indicating that we can apply L'Hôpital's Rule to evaluate the limit.

Step 2: Applying L'Hôpital's Rule

Using L'Hôpital's Rule, we differentiate the numerator and denominator:

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

Denominator: \( \frac{d}{dx}[(\tan(2x))^3] \)

Calculating these derivatives can be complex, but we can simplify the analysis by recognizing that as \( x \) approaches 0, both the numerator and denominator will still yield a limit that we can evaluate.

Finding the Limit Directly

Instead of differentiating, we can analyze the limit directly:

Using the Taylor series expansion around \( x = 0 \):

• \( \sqrt{x^2 + 1} \approx 1 + \frac{x^2}{2} \)
• \( \log(1 + x) \approx x \)
• \( \tan(2x) \approx 2x \)

Substituting these approximations into our limit gives:

\( \lim_{x \to 0} \frac{(\frac{x^2}{2})(\frac{x}{2})}{(2x)^3} = \lim_{x \to 0} \frac{\frac{x^3}{4}}{8x^3} = \lim_{x \to 0} \frac{1}{32} = \frac{1}{32} \)

Step 3: Comparing the Limit to F(0)

Now we compare the limit we found to the value of the function at \( x = 0 \):

We have:

• Limit as \( x \to 0 \): \( \frac{1}{32} \)
• Value at \( x = 0 \): \( F(0) = 16 \)

Since \( \frac{1}{32} \neq 16 \), the function is discontinuous at \( x = 0 \).

Identifying the Type of Discontinuity

This type of discontinuity is classified as a jump discontinuity because the limit does not equal the function value at that point.

Redefining the Function for Continuity

To make the function continuous at \( x = 0 \), we can redefine \( F(0) \) to match the limit:

Thus, we can redefine:

Let \( F(0) = \frac{1}{32} \) instead of 16.

With this adjustment, the function would be continuous at \( x = 0 \) since the limit and the function value would now be equal.