To find the limit

$\lim _{x\to 0}\left(\frac{x^2-a^2}{x+a}\right),$

let’s approach the problem step by step:

1. Substitute $x = 0$ directly into the expression:

   $\frac{0^2 - a^2}{0 + a} = \frac{-a^2}{a} = -a.$

Thus, the limit is:

$\boxed{-a}.$

This is the solution for any real number $a$, except for $a = 0$, which we will handle separately.

2. Special case $a = 0$:

   Substituting $a = 0$ into the original expression:

   $\lim_{x \to 0} \left( \frac{x^2 - 0^2}{x + 0} \right) = \lim_{x \to 0} \frac{x^2}{x} = \lim_{x \to 0} x = 0.$

So, for $a = 0$, the limit is:

$\boxed{0}.$

Final Answer:

• For $a \neq 0$, the limit is $\boxed{-a}$.
• For $a = 0$, the limit is $\boxed{0}$.

Would you like more details or have any questions?

Here are 5 related questions:

1. What happens if we apply L'Hopital's Rule in this problem?
2. How would this limit change if $a$ was restricted to positive values?
3. Can we generalize this type of limit for other forms of rational functions?
4. What if the denominator was $x - a$ instead of $x + a$?
5. How can we graphically interpret the result of this limit?

Tip: When approaching limits, always check if direct substitution is possible before considering alternative methods like factoring or L'Hopital's Rule.