SL 5.1—Introduction of differential calculus

Substituting $x=3$ gives $\frac{9-9}{3-3}=\frac{0}{0}$, an indeterminate form. Direct substitution fails, so we must simplify first.

The numerator factors as $x^2-9=(x-3)(x+3)$. So: $\frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3(x\ne 3)$

Now we take the limit: ${\lim }_{x\to 3}(x+3)=3+3=6$ The limit equals $6$, even though $f(3)$ is undefined.

The correct answer is: The limit equals $6$ because $\frac{x^2-9}{x-3}=x+3$ for $x\ne 3$.

We need to find $\underset{x\to 3}{\lim }\frac{x^2-9}{x-3}$. The key is that a limit describes what the function approaches, not its value at the point.

The option 'limit does not exist because $f(3)$ is undefined' is wrong. A function being undefined at a point does not prevent the limit from existing — limits are about behaviour near the point.

The option 'limit equals $0$' confuses the numerator's value at $x=3$ with the limit. After cancelling the common factor $(x-3)$, the expression simplifies to $x+3$, which approaches $6$, not $0$.

The option 'limit equals $3$' is incorrect. There is no valid reason the limit would equal $3$; this appears to misread the denominator as the result.

The remaining option — the limit equals $6$ — is correct, confirmed by factorising: $\frac{(x-3)(x+3)}{x-3}=x+3\to 6$ as $x\to 3$.

As $x$ approaches $4$ from below: $h(3.9)=7.8$, $h(3.99)=7.98$. These values are getting closer and closer to $8$.

As $x$ approaches $4$ from above: $h(4.01)=8.02$, $h(4.1)=8.2$. These values are also getting closer and closer to $8$.

Both the left-hand and right-hand approaches give $h(x)\to 8$. Since both sides agree, the limit exists and equals $8$.

${\lim }_{x\to 4}h(x)=8$

We must estimate the limit from a table of values by observing the trend from both sides of $x=4$.

$7.98$ is just one data point in the table (at $x=3.99$), not the value the function is approaching. Selecting a single table entry is not the same as finding the limit.

$8.02$ is similarly just the value at $x=4.01$. As $x$ gets even closer to $4$, $h(x)$ would approach $8$, not stay at $8.02$.

The claim that the limit does not exist because $h(4)$ is not shown is incorrect. A limit only requires observing what the function approaches — the actual value at $x=4$ is irrelevant to the limit.

From both sides, $h(x)\to 8$. The correct estimate is $\underset{x\to 4}{\lim }h(x)=8$.