The image contains the following limit problems:

1. $\lim_{{x \to 10}} \frac{{x - 10}}{{x - 10}}$
2. $\lim_{{x \to -30}} \frac{{36 - x^2}}{{x^2 - 30}}$

Problem 1:

$\lim_{{x \to 10}} \frac{{x - 10}}{{x - 10}}$ This is a standard limit problem where the numerator and denominator are identical. We can simplify the expression: $\frac{{x - 10}}{{x - 10}} = 1 \quad \text{(for all } x \neq 10\text{)}.$ Thus, the limit becomes: $\lim_{{x \to 10}} 1 = 1$ So, the limit is 1.

Problem 2:

$\lim_{{x \to -30}} \frac{{36 - x^2}}{{x^2 - 30}}$ We need to substitute $x = -30$ into the expression. If we do that directly, we get: $36 - (-30)^2 = 36 - 900 = -864$ and $(-30)^2 - 30 = 900 - 30 = 870.$ Thus, the expression becomes: $\frac{{-864}}{{870}}.$ Simplifying this fraction: $\frac{{-864}}{{870}} = -\frac{{864}}{{870}} = -\frac{{432}}{{435}}.$ This is the simplified form of the limit, which equals approximately $-0.9931$.

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Related questions:

1. What are the conditions for limits to exist?
2. How do we deal with indeterminate forms when calculating limits?
3. What is the concept of one-sided limits?
4. How can L'Hopital's Rule be applied to limits of indeterminate forms?
5. How does simplifying rational functions help in solving limits?

Tip:

Always check if direct substitution gives a determinate value or an indeterminate form like $\frac{0}{0}$. If indeterminate, explore methods like factoring, rationalization, or L'Hopital's Rule.