SL 5.1—Introduction to Limits

Direct substitution of $x=3$ gives $\frac{0}{0}$, so the expression must be simplified algebraically first.

$x^2-9=(x-3)(x+3)$, so $\frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3$ for $x\ne 3$.

Since the limit only asks about behaviour near $x=3$, substitute into $x+3$: $3+3=6$.

$\underset{x\to 3}{\lim }\frac{x^2-9}{x-3}=6$, even though the original function is undefined at $x=3$.

$0$ would result from wrongly assuming the whole expression is $0/0=0$, ignoring the need to simplify.

Undefined confuses the function value $f(3)$ with the limit; the limit can exist even when $f(3)$ does not.

$3$ comes from mistakenly substituting $x=3$ into only part of the factored expression, such as forgetting to add 3.

Only $6$ correctly reflects the simplified limit calculation.

The open circle shows the $y$-value the curve approaches, while the solid dot shows the actual defined function value $f(5)$.

Tracing the curve from the left and right of $x=5$, both sides approach the open circle at $y=8$.

Since left and right approaches agree at $y=8$, the limit exists and equals 8.

${\lim }_{x\to 5}f(x)=8$, even though $f(5)=3$ is the actual function value at that point.

$3$ is $f(5)$, the plotted (solid) point value — a common mistake of reading the function value instead of the approached value.

The limit does exist here because both sides approach the same $y$-value; 'does not exist' would only apply if left and right disagreed.

$5$ confuses the $x$-value being approached with the limit's $y$-value output.

The correct limit value is the open circle's $y$-coordinate, $8$.