Limit and Its Relation to Differentiation

2.1 Limit and Its Relation to Differentiation
 
The image is an educational diagram explaining the concept of limits in mathematics. At the top, there is the title ‘LIMIT’ with the Pandai logo above it. - To the left, there is a box labeled ‘Gradient of Tangent’ connected to the title by an arrow labeled ‘application’. - To the right, there is a box with the notation ‘lim x→0 f(x)’ connected to the title by an arrow labeled ‘notation’. - Below the title, there is a formula ‘lim Δx→0 Δy/Δx’ in a box, with ‘Δy’ and ‘Δx’ labeled as ‘read as 'delta x' (a small change in x)’ and ‘read as 'delta y.
 
Definition of Limit
When \(x\) approaches \(a\), where \(x\neq a\), the limit for \(f(x)\) is \(L\) can be written as \(\lim_{x \to a} f(x) = L\).
 
Methods to Determine the Limit

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

Condition Method
\(f(a) \neq \dfrac{0}{0}\) The value of \(\lim_{x \to a} f(x)\) can be obtained, that is \(\lim_{x \to a} f(x) = f(a)\).
\(f(a) = \dfrac{0}{0}\)

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

  • Factorisation,
  • Rationalising the numerator or denominator of the function.
 
Definition of Tangent
A tangent to a curve at a point is a straight line that touches the curve at only that point.
 
Gradient of Tangent
Figure

A graph illustrating a function of y, featuring a tangent line and a specific point marked for reference.

Formula

Based on the graph above, the line \(AT\) is a tangent to the curve \(y=x^2\) at the point \(A\):

\(\text{Gradient of tangent }AT = \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}}\).

 
Determining Gradients Using Limit
Figure

A graph displaying the function of y alongside its tangent line, highlighting the path of a line in a mathematical context.

Description
  • Based on the graph above, the gradient of the line \(BC\) can be calculated as follows:

\(\begin{aligned} &\text{Gradient of the line }BC \\&= \dfrac{CD}{BD}\\ &= \dfrac{(y- \delta y) -y}{(x+ \delta x) -x}\\ &= \dfrac{\delta y}{\delta x}. \end{aligned}\)

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

\(\begin{aligned} &\text{Gradient of the curve at }B\\ &= \text{Gradient of tangent }BT\\ &= \lim_{\delta x \to 0} \dfrac{\delta y}{\delta x}. \end{aligned}\)

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

\(\begin{aligned} \ \dfrac{dy}{dx} &= \lim_{\delta x \to 0} \dfrac{\delta y}{\delta x}\\\\ &= \lim_{\delta x \to 0} \dfrac{f(x+\delta x) - f(x)}{\delta x}. \end{aligned}\)

 
Example \(1\)
Question

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.

\(\begin{aligned} \lim_{x \to 1} \dfrac{x^2- 1}{x-1} &= \lim_{x \to 1} \dfrac{(x+1)(x-1)}{x-1} \\\\ &= \lim_{x \to 1} \ (x+1)\\\\ &= 1+1\\\ &=2. \end{aligned}\)

 
Example \(2\)
Question

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
 
The image is an educational diagram explaining the concept of limits in mathematics. At the top, there is the title ‘LIMIT’ with the Pandai logo above it. - To the left, there is a box labeled ‘Gradient of Tangent’ connected to the title by an arrow labeled ‘application’. - To the right, there is a box with the notation ‘lim x→0 f(x)’ connected to the title by an arrow labeled ‘notation’. - Below the title, there is a formula ‘lim Δx→0 Δy/Δx’ in a box, with ‘Δy’ and ‘Δx’ labeled as ‘read as 'delta x' (a small change in x)’ and ‘read as 'delta y.
 
Definition of Limit
When \(x\) approaches \(a\), where \(x\neq a\), the limit for \(f(x)\) is \(L\) can be written as \(\lim_{x \to a} f(x) = L\).
 
Methods to Determine the Limit

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

Condition Method
\(f(a) \neq \dfrac{0}{0}\) The value of \(\lim_{x \to a} f(x)\) can be obtained, that is \(\lim_{x \to a} f(x) = f(a)\).
\(f(a) = \dfrac{0}{0}\)

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

  • Factorisation,
  • Rationalising the numerator or denominator of the function.
 
Definition of Tangent
A tangent to a curve at a point is a straight line that touches the curve at only that point.
 
Gradient of Tangent
Figure

A graph illustrating a function of y, featuring a tangent line and a specific point marked for reference.

Formula

Based on the graph above, the line \(AT\) is a tangent to the curve \(y=x^2\) at the point \(A\):

\(\text{Gradient of tangent }AT = \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}}\).

 
Determining Gradients Using Limit
Figure

A graph displaying the function of y alongside its tangent line, highlighting the path of a line in a mathematical context.

Description
  • Based on the graph above, the gradient of the line \(BC\) can be calculated as follows:

\(\begin{aligned} &\text{Gradient of the line }BC \\&= \dfrac{CD}{BD}\\ &= \dfrac{(y- \delta y) -y}{(x+ \delta x) -x}\\ &= \dfrac{\delta y}{\delta x}. \end{aligned}\)

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

\(\begin{aligned} &\text{Gradient of the curve at }B\\ &= \text{Gradient of tangent }BT\\ &= \lim_{\delta x \to 0} \dfrac{\delta y}{\delta x}. \end{aligned}\)

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

\(\begin{aligned} \ \dfrac{dy}{dx} &= \lim_{\delta x \to 0} \dfrac{\delta y}{\delta x}\\\\ &= \lim_{\delta x \to 0} \dfrac{f(x+\delta x) - f(x)}{\delta x}. \end{aligned}\)

 
Example \(1\)
Question

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.

\(\begin{aligned} \lim_{x \to 1} \dfrac{x^2- 1}{x-1} &= \lim_{x \to 1} \dfrac{(x+1)(x-1)}{x-1} \\\\ &= \lim_{x \to 1} \ (x+1)\\\\ &= 1+1\\\ &=2. \end{aligned}\)

 
Example \(2\)
Question

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,

\(\begin{aligned} \dfrac{dy}{dx} &= \lim_{\delta x \to 0} \dfrac{\delta y}{\delta x}\\\\ &= \lim_{\delta x \to 0} (6x+3\delta x)\\\\ &=6x+3(0)\\ &= 6x. \end{aligned}\)