Limit and Its Relation to Differentiation

 2.1 Limit and Its Relation to Differentiation

• When $$x$$ approaches $$a$$, where $$x \neq a$$ , the limit for $$f(x)$$ is $$L$$ can be written as
 $$\ \lim_{x \to a} f(x) = L \$$

• The steps to determine $$\lim_{x \to a} f(x)$$ where $$a \in ℝ$$ are as follows:

To find the limit value of a function $$f(x)$$, we substitute $$x=a$$ directly into the function $$f(x)$$.

(a)
 $$\ f(a) \neq \dfrac{0}{0} \$$

The value of $$\lim_{x \to a} f(x)$$ can be obtained, that is $$\lim_{x \to a} f(x) = f(a)$$

(b)
 $$\ f(a) = \dfrac{0}{0} \$$

Determine $$\lim_{x \to a} f(x)$$ by using the following methods:

• Factorisation
• Rationalising the numerator or denominator of the function

Example

Determine the limit value for each of the following functions:

 (a) $$\lim_{x \to 4} \dfrac{3- \sqrt{x}}{x+2}$$ (b) $$\lim_{x \to 1} \dfrac{x^2- 1}{x-1}$$ Solution: (a) Use direct substitution. \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} (b) When $$x=1$$, $$\lim_{x \to 1} \dfrac{x^2- 1}{x-1}$$ is in the indeterminate form, $$\dfrac{0}{0}$$. Thus, we need to factorise and eliminate the common factor before we can use direct substitution. \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}

• A tangent to a curve at a point is a straight line that touches the curve at only that point

• Based on the graph above, the line $$AT$$ is a tangent to the curve $$y=x^2$$ at the point $$A$$:
 \begin{aligned}\\ \ \text{Gradient of tangent }AT = \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}} \ \\\\ \end{aligned}

• Other methods to find the gradient of the curve at a particular point, that is by using the idea of limits

• 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}

• For the curve $$y=f(x)$$, the gradient function of the tangent at any point can be obtained using the following formulae:

 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}

• $$\lim_{\delta x \to 0} \dfrac{\delta y}{\delta x}$$ 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 $$(x, f(x))$$

 \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}

Contoh

Find $$\dfrac{dy}{dx}$$ by using first principle for each of the following functions $$y=f(x)$$.

 (a) $$y=3x$$ (b) $$y=3x^2$$ Solution: (a) Given $$y=f(x)=3x$$. \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} Hence, \begin{aligned} \dfrac{dy}{dx} &= \lim_{\delta x \to 0} \dfrac{\delta y}{\delta x}\\\\ &= \lim_{\delta x \to 0} 3\\\\ &=3 \end{aligned} (b) Given $$y=f(x)=3x^2$$. \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} Hence, \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}

Limit and Its Relation to Differentiation

 2.1 Limit and Its Relation to Differentiation

• When $$x$$ approaches $$a$$, where $$x \neq a$$ , the limit for $$f(x)$$ is $$L$$ can be written as
 $$\ \lim_{x \to a} f(x) = L \$$

• The steps to determine $$\lim_{x \to a} f(x)$$ where $$a \in ℝ$$ are as follows:

To find the limit value of a function $$f(x)$$, we substitute $$x=a$$ directly into the function $$f(x)$$.

(a)
 $$\ f(a) \neq \dfrac{0}{0} \$$

The value of $$\lim_{x \to a} f(x)$$ can be obtained, that is $$\lim_{x \to a} f(x) = f(a)$$

(b)
 $$\ f(a) = \dfrac{0}{0} \$$

Determine $$\lim_{x \to a} f(x)$$ by using the following methods:

• Factorisation
• Rationalising the numerator or denominator of the function

Example

Determine the limit value for each of the following functions:

 (a) $$\lim_{x \to 4} \dfrac{3- \sqrt{x}}{x+2}$$ (b) $$\lim_{x \to 1} \dfrac{x^2- 1}{x-1}$$ Solution: (a) Use direct substitution. \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} (b) When $$x=1$$, $$\lim_{x \to 1} \dfrac{x^2- 1}{x-1}$$ is in the indeterminate form, $$\dfrac{0}{0}$$. Thus, we need to factorise and eliminate the common factor before we can use direct substitution. \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}

• A tangent to a curve at a point is a straight line that touches the curve at only that point

• Based on the graph above, the line $$AT$$ is a tangent to the curve $$y=x^2$$ at the point $$A$$:
 \begin{aligned}\\ \ \text{Gradient of tangent }AT = \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}} \ \\\\ \end{aligned}

• Other methods to find the gradient of the curve at a particular point, that is by using the idea of limits

• 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}

• For the curve $$y=f(x)$$, the gradient function of the tangent at any point can be obtained using the following formulae:

 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}

• $$\lim_{\delta x \to 0} \dfrac{\delta y}{\delta x}$$ 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 $$(x, f(x))$$

 \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}

Contoh

Find $$\dfrac{dy}{dx}$$ by using first principle for each of the following functions $$y=f(x)$$.

 (a) $$y=3x$$ (b) $$y=3x^2$$ Solution: (a) Given $$y=f(x)=3x$$. \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} Hence, \begin{aligned} \dfrac{dy}{dx} &= \lim_{\delta x \to 0} \dfrac{\delta y}{\delta x}\\\\ &= \lim_{\delta x \to 0} 3\\\\ &=3 \end{aligned} (b) Given $$y=f(x)=3x^2$$. \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} Hence, \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}