Question

$\lim_{h \to 0} \frac{\frac{4}{x+3h} - \frac{4}{x}}{h}$
answer: $\boxed{}$

Explanation:

Step1: Simplify the numerator

First, we simplify the numerator \(\frac{4}{x + 3h}-\frac{4}{x}\). Find a common denominator, which is \(x(x + 3h)\). Then we have:
\[

$$\begin{align*} \frac{4}{x + 3h}-\frac{4}{x}&=\frac{4x-4(x + 3h)}{x(x + 3h)}\\ &=\frac{4x-4x-12h}{x(x + 3h)}\\ &=\frac{-12h}{x(x + 3h)} \end{align*}$$

\]

Step2: Substitute back into the limit

Now the original limit \(\lim_{h
ightarrow0}\frac{\frac{4}{x + 3h}-\frac{4}{x}}{h}\) becomes \(\lim_{h
ightarrow0}\frac{\frac{-12h}{x(x + 3h)}}{h}\). Dividing by \(h\) is the same as multiplying by \(\frac{1}{h}\), so we can rewrite this as:
\[
\lim_{h
ightarrow0}\frac{-12h}{x(x + 3h)\cdot h}
\]

Step3: Cancel out \(h\)

We can cancel out the \(h\) terms (assuming \(h
eq0\), which is valid since we are taking the limit as \(h
ightarrow0\) and not evaluating at \(h = 0\)):
\[
\lim_{h
ightarrow0}\frac{-12}{x(x + 3h)}
\]

Step4: Evaluate the limit as \(h

ightarrow0\)
Now we substitute \(h = 0\) into the expression:
\[
\frac{-12}{x(x+3\cdot0)}=\frac{-12}{x\cdot x}=-\frac{12}{x^{2}}
\]

Answer:

\(-\dfrac{12}{x^{2}}\)