Question

this question is designed to be answered without a calculator.
if $f(x) = 5(x^2 - 1)$, then $lim_{x \to 2} \frac{f(x) - f(2)}{x - 2} =$
\\(\circ\\) 0.
\\(\circ\\) 5.
\\(\circ\\) 10.
\\(\circ\\) 20.

Explanation:

Step1: Recall the definition of derivative

The limit \(\lim_{x\to a}\frac{f(x)-f(a)}{x - a}\) is the definition of the derivative of \(f(x)\) at \(x=a\), i.e., \(f^{\prime}(a)\). So here we can find \(f^{\prime}(x)\) first and then evaluate it at \(x = 2\), or we can simplify the expression \(\frac{f(x)-f(2)}{x - 2}\) directly. Let's try simplifying directly.

First, find \(f(2)\):
Given \(f(x)=5(x^{2}-1)\), substitute \(x = 2\) into \(f(x)\):
\(f(2)=5(2^{2}-1)=5(4 - 1)=5\times3 = 15\)

Then, substitute \(f(x)\) and \(f(2)\) into \(\frac{f(x)-f(2)}{x - 2}\):
\(\frac{5(x^{2}-1)-15}{x - 2}=\frac{5x^{2}-5 - 15}{x - 2}=\frac{5x^{2}-20}{x - 2}\)

Step2: Factor the numerator

Factor out 5 from the numerator: \(\frac{5(x^{2}-4)}{x - 2}\)
Notice that \(x^{2}-4\) is a difference of squares, so \(x^{2}-4=(x - 2)(x + 2)\)
Substitute this into the expression: \(\frac{5(x - 2)(x + 2)}{x - 2}\)

Step3: Cancel the common factor

As long as \(x
eq2\) (which is true when taking the limit as \(x\to2\), we are approaching 2 but not equal to 2), we can cancel the \((x - 2)\) terms:
\(5(x + 2)\)

Step4: Evaluate the limit

Now, take the limit as \(x\to2\) of \(5(x + 2)\):
\(\lim_{x\to2}5(x + 2)=5(2 + 2)=5\times4 = 20\)? Wait, no, wait, let's check again. Wait, no, wait, maybe we made a mistake in differentiation approach. Wait, let's use the derivative approach. The derivative of \(f(x)=5(x^{2}-1)=5x^{2}-5\). The derivative \(f^{\prime}(x)=10x\). Then \(f^{\prime}(2)=10\times2 = 20\)? Wait, but wait, when we simplified \(\frac{5(x^{2}-4)}{x - 2}=\frac{5(x - 2)(x + 2)}{x - 2}=5(x + 2)\) (for \(x
eq2\)), then \(\lim_{x\to2}5(x + 2)=5(2 + 2)=20\)? But wait, let's check the arithmetic again. Wait, \(f(x)=5(x^{2}-1)\), so \(f(x)-f(2)=5x^{2}-5-(5(4 - 1))=5x^{2}-5 - 15=5x^{2}-20=5(x^{2}-4)=5(x - 2)(x + 2)\). Then \(\frac{f(x)-f(2)}{x - 2}=\frac{5(x - 2)(x + 2)}{x - 2}=5(x + 2)\) (x≠2). Then as \(x\to2\), \(5(x + 2)\to5(2 + 2)=20\)? But wait, the options have 20 as an option. But wait, let's check with the derivative. \(f^{\prime}(x)=10x\), so \(f^{\prime}(2)=20\). So that's correct. Wait, but earlier when I thought maybe mistake, but no. Wait, but let's check the initial simplification again. \(f(x)-f(2)=5(x^{2}-1)-5(2^{2}-1)=5(x^{2}-1)-5(3)=5x^{2}-5 - 15=5x^{2}-20\). Then \(\frac{5x^{2}-20}{x - 2}=\frac{5(x^{2}-4)}{x - 2}=\frac{5(x - 2)(x + 2)}{x - 2}=5(x + 2)\) (x≠2). Then limit as x approaches 2 is 5(4)=20. So the answer is 20? But wait, the options include 20. Wait, but let's check the derivative again. \(f(x)=5x^{2}-5\), \(f^{\prime}(x)=10x\), so at x=2, \(f^{\prime}(2)=20\). So that's correct.

Wait, but wait, the problem is \(\lim_{x\to2}\frac{f(x)-f(2)}{x - 2}\), which is the definition of the derivative at x=2. So derivative of f at 2 is 20. So the answer is 20.

Answer:

20