The problem given in the image is:

$\lim_{{x \to 3}} \left(\frac{1}{x+1} - \frac{1}{4}\right)$

Let's solve this step by step.

Step 1: Plug in the value of $x = 3$

We first substitute $x = 3$ directly into the expression to see if we can evaluate the limit directly.

$\frac{1}{x+1} - \frac{1}{4} = \frac{1}{3+1} - \frac{1}{4} = \frac{1}{4} - \frac{1}{4} = 0$

So, when $x = 3$, the expression simplifies to 0.

Step 2: Conclusion

Since the direct substitution of $x = 3$ gives us a defined value (0), the limit exists, and we can conclude that:

$\lim_{{x \to 3}} \left(\frac{1}{x+1} - \frac{1}{4}\right) = 0$

Final Answer

$\boxed{0}$

This is the correct and final answer.

Do you have any questions or need more details on any step?

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Here are 5 related questions for further practice:

1. What is the limit of $\lim_{{x \to 2}} \left(\frac{1}{x-2} + 3\right)$?
2. Evaluate $\lim_{{x \to 1}} \left(\frac{2x+1}{x-1}\right)$.
3. Determine the limit $\lim_{{x \to 0}} \frac{\sin(x)}{x}$.
4. Calculate $\lim_{{x \to 0}} \left(\frac{1}{x} + 2\right)$.
5. Solve $\lim_{{x \to -1}} \left(\frac{x^2-1}{x+1}\right)$.

Tip: When finding limits, always check if the function is continuous at the point by substituting the value directly. If it results in an indeterminate form, consider algebraic manipulation or L'Hôpital's rule.