Answer is: option3

Solution:

Substituting \( z = 0 \):

Numerator:

\[ \sin(2(0)) + 7(0)^2 - 2(0) = 0 + 0 - 0 = 0 \]

Denominator:

\[ (0)^2 (0 + 1)^2 = 0^2 \cdot 1^2 = 0 \]

Since we get the indeterminate form \( \frac{0}{0} \), we apply L'Hôpital’s Rule by differentiating the numerator and denominator separately.

\[ \frac{d}{dz} (\sin(2z) + 7z^2 - 2z) = \cos(2z) \cdot 2 + 14z - 2 \] \[ = 2\cos(2z) + 14z - 2 \]

\[ \frac{d}{dz} (z^2 (z + 1)^2) \]

Using the product rule where \( u = z^2 \) and \( v = (z + 1)^2 \),

\[ \frac{d}{dz} (z^2) = 2z, \quad \frac{d}{dz} ((z + 1)^2) = 2(z + 1) \] \[ \frac{d}{dz} (z^2 (z + 1)^2) = 2z (z + 1)^2 + z^2 \cdot 2(z + 1) \] \[ = 2z(z + 1)^2 + 2z^2 (z + 1) \] \[ = 2(z + 1)(z + 1 + z) \] \[ = 2z (z + 1)(2z + 1) \]

Substituting \( z = 0 \) in the new fraction:

Numerator:

\[ 2\cos(0) + 14(0) - 2 = 2(1) + 0 - 2 = 0 \]

Denominator:

\[ 2(0)(0 + 1)(2(0) + 1) = 0 \]

Since we still have \( \frac{0}{0} \), we apply L'Hôpital’s Rule again.

Second Derivative of the Numerator:

\[ \frac{d}{dz} (2\cos(2z) + 14z - 2) = -4\sin(2z) + 14 \]

Second Derivative of the Denominator:

Using the product rule,

\[ \frac{d}{dz} (2z (z + 1)(2z + 1)) \]

\[ = 2 (z + 1)(2z + 1) + 2z \frac{d}{dz} ((z + 1)(2z + 1)) \]

\[ = 2 (z + 1)(2z + 1) + 2[(z + 1) \cdot 2 + 2z + 1] \]

\[ = 2 (z + 1)(2z + 1) + 2(4z + 3) \]

Evaluating at \( z = 0 \):

Numerator:

\[ -4\sin(0) + 14 = 0 + 14 = 14 \]

Denominator:

\[ 2(0 + 1)(2(0) + 1) + 2(0)(4(0) + 3) = 2(1)(1) + 0 = 2 \]

Thus, the limit is:

\[ \frac{14}{2} = 7 \]