Question

If $f^{\prime}(5)=\frac{3}{5}$ then the value of $\lim_{h\rightarrow 0}\frac{f(5+10h)-f(5)}{h}$ is equal to

• A. 1
• B. 2
• C. 3
• D. 5
• E. 6

Hint:

Logic Tip: A very useful shortcut for limits resembling the derivative definition is: $\lim_{h \to 0} \frac{f(x+ah)-f(x)}{h} = a \cdot f'(x)$. Here, $a=10$, so the answer is immediately $10 \cdot f'(5) = 10(3/5) = 6$.

Answer 1

Answer: (E)

Concept:
The standard definition of a derivative at a point $x=a$ is: $$f'(a) = \lim_{t\rightarrow 0}\frac{f(a+t)-f(a)}{t}$$ To evaluate limits that look similar to this definition, we must algebraically manipulate the expression to perfectly match the formula.
Step 1: Identify the components and set up a substitution.
We are given the expression: $$\lim_{h\rightarrow 0}\frac{f(5+10h)-f(5)}{h}$$ Notice that the term added to 5 inside the function is $10h$, but the denominator is only $h$. Let $t = 10h$. As $h \rightarrow 0$, $t \rightarrow 0$. Also, we can express $h$ in terms of $t$: $h = \frac{t}{10}$.
Step 2: Substitute and rearrange the limit.
Replace $10h$ and $h$ in the limit: $$= \lim_{t\rightarrow 0}\frac{f(5+t)-f(5)}{\frac{t}{10}}$$ Multiply the numerator and denominator by 10 to clear the complex fraction: $$= \lim_{t\rightarrow 0} 10 \cdot \frac{f(5+t)-f(5)}{t}$$ Since limits allow constant multiples to be factored out: $$= 10 \cdot \left( \lim_{t\rightarrow 0} \frac{f(5+t)-f(5)}{t} \right)$$
Step 3: Apply the definition of the derivative and evaluate.
The expression inside the parenthesis is exactly the definition of $f'(5)$. $$= 10 \cdot f'(5)$$ We are given that $f'(5) = \frac{3}{5}$. Substitute this value: $$= 10 \cdot \left(\frac{3}{5}\right)$$ $$= \frac{30}{5} = 6$$