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- When \(x\) approaches \(a\), where \(x \neq a\) , the limit for \(f(x)\) is \(L\) can be written as
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\(\ \lim_{x \to a} f(x) = L \ \) |
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- The steps to determine \(\lim_{x \to a} f(x)\) where \(a \in ℝ\) are as follows:
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To find the limit value of a function \(f(x)\), we substitute \(x=a\) directly into the function \(f(x)\).
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(a) |
\(\ f(a) \neq \dfrac{0}{0} \ \) |
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The value of \(\lim_{x \to a} f(x)\) can be obtained, that is \(\lim_{x \to a} f(x) = f(a)\)
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(b) |
\(\ f(a) = \dfrac{0}{0} \ \) |
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Determine \(\lim_{x \to a} f(x)\) by using the following methods:
- Factorisation
- Rationalising the numerator or denominator of the function
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Example |
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Determine the limit value for each of the following functions: |
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(a) |
\(\lim_{x \to 4} \dfrac{3- \sqrt{x}}{x+2}\) |
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(b) |
\(\lim_{x \to 1} \dfrac{x^2- 1}{x-1}\) |
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Solution: |
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(a) |
Use direct substitution. |
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\(\begin{aligned} \lim_{x \to 4} \dfrac{3- \sqrt{x}}{x+2} &= \dfrac{3- \sqrt{4}}{4+2} \\\\ &= \dfrac{3-2}{4+2}\\\\ &= \dfrac{1}{6} \end{aligned}\)
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(b) |
When \(x=1\), \(\lim_{x \to 1} \dfrac{x^2- 1}{x-1}\) is in the indeterminate form, \(\dfrac{0}{0}\). |
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Thus, we need to factorise and eliminate the common factor before we can use direct substitution.
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\(\begin{aligned} \lim_{x \to 1} \dfrac{x^2- 1}{x-1} &= \lim_{x \to 1} \dfrac{(x+1)(x-1)}{x-1} \\\\ &= \lim_{x \to 1} \ (x+1)\\\\ &= 1+1\\\ &=2 \end{aligned}\)
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- A tangent to a curve at a point is a straight line that touches the curve at only that point
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- Based on the graph above, the line \(AT\) is a tangent to the curve \(y=x^2\) at the point \(A\):
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\(\begin{aligned}\\ \ \text{Gradient of tangent }AT = \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}} \ \\\\ \end{aligned}\) |
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- Other methods to find the gradient of the curve at a particular point, that is by using the idea of limits
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- Based on the graph above, the gradient of the line \(BC\) can be calculated as follows:
Gradient of the line \(BC\)
\(\begin{aligned} &= \dfrac{CD}{BD}\\ &= \dfrac{(y- \delta y) -y}{(x+ \delta x) -x}\\ &= \dfrac{\delta y}{\delta x} \end{aligned}\)
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- For the curve \(y=f(x)\), the gradient function of the tangent at any point can be obtained using the following formulae:
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Gradient of the curve at \(B\)
\(\begin{aligned} &= \text{Gradient of tangent }BT\\ &= \lim_{\delta x \to 0} \dfrac{\delta y}{\delta x} \end{aligned}\)
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- \( \lim_{\delta x \to 0} \dfrac{\delta y}{\delta x}\) is the first derivatives of the function and is written with the symbol \(\dfrac{dy}{dx}\)
- The gradient function \(\dfrac{dy}{dx}\) is known as differentiation using first principles
- It can be use to find the gradient of a tangent of a curve \(y=f(x)\) at a point \((x, f(x))\)
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\(\begin{aligned} \ \dfrac{dy}{dx} &= \lim_{\delta x \to 0} \dfrac{\delta y}{\delta x}\\\\ &= \lim_{\delta x \to 0} \dfrac{f(x+\delta x) - f(x)}{\delta x} \ \end{aligned}\) |
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Contoh |
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Find \(\dfrac{dy}{dx}\) by using first principle for each of the following functions \(y=f(x)\). |
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(a) |
\(y=3x\) |
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(b) |
\(y=3x^2\) |
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Solution: |
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(a) |
Given \(y=f(x)=3x\). |
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\(\begin{aligned} \delta y &= f(x+\delta x) - f(x)\\ &= 3(x+\delta x) - 3x\\ &= 3x + 3 \delta x-3x\\ &= 3 \delta x\\ \dfrac{\delta y}{\delta x}&= 3 \end{aligned}\)
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Hence, |
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\(\begin{aligned} \dfrac{dy}{dx} &= \lim_{\delta x \to 0} \dfrac{\delta y}{\delta x}\\\\ &= \lim_{\delta x \to 0} 3\\\\ &=3 \end{aligned}\)
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(b) |
Given \(y=f(x)=3x^2\). |
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\(\begin{aligned} \delta y &= f(x+\delta x) - f(x)\\ &= 3(x+\delta x)^2 - 3x^2\\ &= 3[x^2 + 2x(\delta x) + (\delta x)^2] - 3x^2\\ &= 3x^2 + 6x(\delta x) + 3(\delta x)^2 - 3x^2\\ &=6x(\delta x) + 3(\delta x)^2\\ \dfrac{\delta y}{\delta x}&= 6x+3\delta x \end{aligned}\)
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Hence, |
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\(\begin{aligned} \dfrac{dy}{dx} &= \lim_{\delta x \to 0} \dfrac{\delta y}{\delta x}\\\\ &= \lim_{\delta x \to 0} (6x+3\delta x)\\\\ &=6x+3(0)\\ &= 6x \end{aligned}\)
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