Evaluate the limit $\lim_{x\to 3}\frac{g(2x+1)-5}{x^2-9}$

Expert intro: This limit is designed to test substitution plus algebraic simplification. The key observation is that as $x\to 3$, the input $2x+1$ approaches $7$, so the value of $g(7)$ matters.

Detailed walkthrough

As $x\to 3$,

$2x+1\to 7.$

So the numerator becomes

$g(2x+1)-5 \to g(7)-5.$

To evaluate the limit in a useful way, we need the given value of $g(7)$. In problems of this form, the intended setup is usually that

$g(7)=5,$

so the numerator approaches $0$ and the expression becomes an indeterminate form $0/0$.

Now factor the denominator:

$x^2-9=(x-3)(x+3).$

If the problem includes a local linearization or derivative condition for $g$ near $7$, you would then use that information to simplify the numerator. For example, if $g$ is differentiable at $7$ and you know $g'(7)$, then

$g(2x+1)\approx g(7)+g'(7)(2x-6).$

Substituting this into the limit gives a derivative-based evaluation.

Without an additional value such as $g(7)$ or $g'(7)$, the limit cannot be determined uniquely from the expression alone. The standard first step is to identify the behavior of the inside function $2x+1$ at $x=3$ and then use the provided information about $g$ at $7$.

💡 Pitfall guide

Do not substitute $x=3$ too early if the numerator also goes to $0$; that can hide an indeterminate form. Also, remember that the inside expression $2x+1$ approaches $7$, not $3$.

🔄 Real-world variant

If the problem states a value such as $g(7)=5$ and also gives $g'(7)$, then the limit is typically solved by rewriting the numerator near $x=3$ as a linear approximation. If instead $g(7)\neq 5$, then the numerator approaches a nonzero constant and the limit usually diverges because $x^2-9\to 0$.

🔍 Related terms

limit evaluation, indeterminate form, linearization