Evaluate lim x→3 (g(2x+1)-5)/(x^2-9)

Key concept: This limit is set up to use a chain-rule style derivative idea. Rewrite the numerator in terms of a new variable and match the denominator near $x=3$ to a standard derivative limit.

Step by step

We want to evaluate

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

Step 1: Substitute the point approached by the inside function

As $x\to 3$,

$2x+1\to 7.$

So the numerator becomes $g(7)-5$. For the limit to be finite, this must be $0$, so we infer

$g(7)=5.$

Step 2: Rewrite the denominator

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

So the limit is

$\lim_{x\to 3}\frac{g(2x+1)-g(7)}{(x-3)(x+3)}.$

Step 3: Make the derivative structure visible

Let

$u=2x+1.$

Then near $x=3$, we have $u\to 7$, and

$u-7=2(x-3).$

So

=\frac{g(u)-g(7)}{u-7}\cdot\frac{u-7}{(x-3)(x+3)}.$$ Since $u-7=2(x-3)$, $$\frac{u-7}{(x-3)(x+3)}=\frac{2}{x+3}.$$ Thus $$\lim_{x\to 3}\frac{g(2x+1)-5}{x^2-9} =\left(\lim_{u\to 7}\frac{g(u)-g(7)}{u-7}\right)\left(\lim_{x\to 3}\frac{2}{x+3}\right).$$ ### Step 4: Finish The first limit is $g'(7)$, and the second is $\frac{2}{6}=\frac13$. So the value is $$\boxed{\frac{1}{3}g'(7)}.$$ If the problem gives a specific value for $g'(7)$, substitute it here. ### Pitfall alert Do not plug in $x=3$ directly before checking whether the numerator is also approaching $0$. The limit is designed to use a derivative form, so the inside point $2x+1\to 7$ matters. ### Try different conditions If the denominator were changed to $2x-6$ instead of $x^2-9$, then the same derivative idea would give a different constant factor, but the core result would still involve $g'(7)$. ### Further reading limit definition, chain rule, derivative at a point