Basic Mathematics · Unit 2
Limits and Continuity
Exam-focused notes for Limits and Continuity (Basic Mathematics, MTH104): what the TU syllabus asks and how it has actually been tested, with 9 solved past questions from this unit.
What this unit covers
- Limit definition and notation
- Limit evaluation techniques
- Limits at infinity
- Infinite limits and vertical asymptotes
- Continuity definition
- Discontinuity types
- Continuous extension of functions
- Epsilon-delta definition
Limit definition and notation
Explanation of Limits and Function Analysis
The statement $\lim{x \to 2} f(x) = 5$ means that as $x$ gets arbitrarily close to $2$ (approaching from both the left and the right), the function values $f(x)$ get arbitrarily close to $5$, irrespective of the value of $f$ at $x = 2$. Formal ($\epsilon$-$...
Full solved answer →Continuity definition
Test whether the function $f(x) = \begin{cases} \dfrac{x^2 - 2x}{x - 2}, & x \ne 2 \ 1, & x = 2 \end{cases}$ is continuous or discontinuous at $x = 2$. Explain. [5]
$$f(x) = \begin{cases} \dfrac{x^2 - 2x}{x - 2}, & x \ne 2 \\[2mm] 1, & x = 2 \end{cases}$$ Point to test: $x = 2$. A function $f$ is continuous at $x = a$ if: 1. $f(a)$ is defined, 2. $\lim{x \to a} f(x)$ exists, 3. $\lim{x \to a} f(x) = f(a)$. $$f(2) = 1 \...
Full solved answer →(question text pending review) # Solution
Main Question [5 marks]
Show that $f(x) = 1 - \sqrt{1 - x^2}$ is continuous on $[-1,1]$
A function is continuous at a point $c$ if $\lim_{x \to
Definition: A function $f$ is continuous on a closed interval $[a,b]$ if it is continuous at every interior point and continuous from the right at $a$ and from the left at $b$. Given: $f(x) = 1 - \sqrt{1 - x^2}$, interval $[-1, 1]$. Step 1: Decompose into e...
Full solved answer →Test whether the function $$f(x) = \begin{cases} \frac{x^2 - 4}{x - 2} & \text{if } x \ne 2 \ 4 & \text{if } x = 2 \end{cases}$$ is continuous or discontinuous at $x = 2$. Explain. [5]
$$f(x) = \begin{cases} \dfrac{x^2 - 4}{x - 2} & \text{if } x \ne 2 \\[2mm] 4 & \text{if } x = 2 \end{cases}$$ Point to test: $x = 2$. A function $f$ is continuous at $x = a$ if: 1. $f(a)$ is defined, 2. $\lim{x \to a} f(x)$ exists, 3. $\lim{x \to a} f(x) = ...
Full solved answer →Limits at infinity
Find the limit of $\lim_{h\to\infty} \frac{\sqrt{6h+25-5}}{h^2}$. [5]
The expression as literally rendered is garbled, but the most sensible interpretation is: $$\lim{h \to \infty} \frac{\sqrt{6h + 25 - 5}}{h^2}$$ - Numerator: $\sqrt{6h + 25 - 5} = \sqrt{6h + 20}$ - Denominator: $h^2$ - Limit variable: $h \to \infty$ Note: Th...
Full solved answer →Discontinuity types
Question
If a function is defined by $$f(x) = \begin{cases} 1 + x & \text{if } x \leq -1 \ x^2 & \text{if } x > -1 \end{cases}$$
Evaluate $f(-3)$, $f(-1)$, and $f(0)$ and sketch the graph. Define different types of discontinuity at a point. At what points the function becomes continuous of the function $$f(x) = \frac{x-2}{x^2-7x+10}$$ [5+5]
Given data: $$f(x) = \begin{cases} 1 + x & \text{if } x \leq -1 \\ x^2 & \text{if } x -1 \end{cases}$$ Points to evaluate: $x = -3,\ -1,\ 0$. Since $-3 \leq -1$, use the first piece $f(x) = 1 + x$: $$f(-3) = 1 + (-3) = \boxed{-2}$$ Since $-1 \leq -1$, use t...
Full solved answer →Epsilon-delta definition
Sketch the graph of the function $f(x) = x^2$
Shifted vertically up to 1 and -2 units and horizontally up to 3 and -2 units.
Find the $\delta$ algebraically for the following functions.
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$\lim_{x \to 5} \sqrt{x-1}$, and $L = 2$, $\epsilon = 1$
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$\lim_{x \to 2} (2x - 2)$, and $L = 6$, $\epsilon = 0.02$
[5+5]
Part 1 (Sketching): Base function $f(x) = x^2$. - Vertical shifts: up 1 unit and down 2 units - Horizontal shifts: right 3 units and left 2 units Part 2 (Finding $\delta$): - Problem 1: $\lim{x \to 5} \sqrt{x-1}$, $L = 2$, $\epsilon = 1$ - Problem 2: $\lim{...
Full solved answer →Continuous extension of functions
Show that $f(x)=\frac{x^2+x-6}{x^2-4}$, $x \neq 2$ has a continuous extension to $x=2$ [5]
- Function: $f(x) = \dfrac{x^2 + x - 6}{x^2 - 4}$, defined for $x \neq 2$. - Target point for extension: $x = 2$. Direct substitution: $$f(2) = \frac{2^2 + 2 - 6}{2^2 - 4} = \frac{0}{0}$$ This is indeterminate, so $f$ is not defined at $x = 2$. We test whet...
Full solved answer →Limit evaluation techniques
Evaluate the following:
$$\lim_{x \to \infty} (x-\sqrt{x^2 + 16})$$
$$\lim_{x \to 1} \left(\frac{\sqrt{6x+10}-5}{x^2}\right)$$
[2.5+2.5]
- Limit 1: $\lim{x \to \infty} (x - \sqrt{x^2 + 16})$ - Limit 2: $\lim{x \to 1} \dfrac{\sqrt{6x+10} - 5}{x^2}$ --- Step 1: Identify the form As $x \to \infty$, this is of the form $\infty - \infty$ (indeterminate). Step 2: Multiply by conjugate $$= \lim{x \...
Full solved answer →Make Unit 2 stick
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