We are tasked with evaluating the limit:

$\lim_{z \to i} \frac{z^2 + 1}{z^6 + 1}$

To find the limit, we'll first substitute $z = i$ into the expression.

Step 1: Substituting $z = i$

Substitute $z = i$ in both the numerator and the denominator.

Numerator:

$z^2 + 1 = i^2 + 1 = (-1) + 1 = 0$

Denominator:

$z^6 + 1 = i^6 + 1$

To evaluate $i^6$, use the property of powers of $i$: $i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1, \quad i^5 = i, \quad i^6 = -1$

Thus, $i^6 + 1 = (-1) + 1 = 0$

Step 2: Indeterminate form

Both the numerator and denominator equal 0 when $z = i$, meaning the limit is in the indeterminate form $\frac{0}{0}$. To resolve this, we'll apply L'Hopital's Rule, which is used to evaluate limits of indeterminate forms.

Step 3: Apply L'Hopital's Rule

L'Hopital's Rule states that for limits of the form $\frac{0}{0}$, we can differentiate the numerator and the denominator and then take the limit.

First, differentiate the numerator and denominator with respect to $z$.

• Derivative of the numerator $z^2 + 1$ with respect to $z$: $\frac{d}{dz}(z^2 + 1) = 2z$

• Derivative of the denominator $z^6 + 1$ with respect to $z$: $\frac{d}{dz}(z^6 + 1) = 6z^5$

Now, the limit becomes: $\lim_{z \to i} \frac{2z}{6z^5}$

Step 4: Simplifying

Simplify the expression: $\lim_{z \to i} \frac{2z}{6z^5} = \lim_{z \to i} \frac{1}{3z^4}$

Step 5: Substitute $z = i$

Now, substitute $z = i$ into the simplified expression: $\frac{1}{3(i)^4} = \frac{1}{3 \cdot 1} = \frac{1}{3}$

Final Answer:

$\lim_{z \to i} \frac{z^2 + 1}{z^6 + 1} = \frac{1}{3}$

Would you like a more detailed explanation of any step?

Here are some related questions to explore:

1. What is L'Hopital's Rule, and when can it be used?
2. How do the powers of $i$ cycle, and why is $i^4 = 1$?
3. Can limits be solved without L'Hopital's Rule in indeterminate forms?
4. How do you evaluate complex limits in multivariable contexts?
5. What are other methods to resolve indeterminate forms like $0/0$?

Tip: Always simplify the expression before applying L'Hopital’s Rule to check if a direct substitution can work.