Mathematics I · Unit 2 · 4 hrs
Limits and Continuity
Exam-focused notes for Limits and Continuity (Mathematics I, MTH117): what the TU syllabus asks and how it has actually been tested, with 8 solved past questions from this unit.
What this unit covers
- Precise definition of Limit
- Limits at infinity
- Continuity
- Horizontal asymptotes
- Vertical and Slant asymptotes
Continuity
Show that the function $f(x) = x^2 + \sqrt{7-x}$ is continuous at $x=4$. [5]
- Function: $f(x) = x^2 + \sqrt{7 - x}$ - Point: $x = 4$ $f(x)$ is continuous at $x = a$ if all three hold: 1. $f(a)$ is defined 2. $\lim{x \to a} f(x)$ exists 3. $\lim{x \to a} f(x) = f(a)$ Here $a = 4$. --- $$f(4) = 4^2 + \sqrt{7 - 4} = 16 + \sqrt{3}$$ Si...
Full solved answer →Define continuity of a function at a point $x=a$. Show that the function $f(x)=\sqrt{1-x^2}$ is continuous on the interval $[-1,1]$. [5]
--- A function f(x) is said to be continuous at a point x = a if the following three conditions are all satisfied: 1. f(a) is defined (the function has a value at x = a) 2. lim(x→a) f(x) exists (the left-hand limit equals the right-hand limit) 3. lim(x→a) f...
Full solved answer →Define continuity on an interval. Show that the function is continuous on the interval $[-1,1]$. $f(x) = 1 - \sqrt{1 - x^2}$ [5]
- Function: $f(x) = 1 - \sqrt{1 - x^2}$ - Interval: $[-1, 1]$ (the problem writes $[1,-1]$, which is the closed interval between $-1$ and $1$) A function $f$ is continuous on a closed interval $[a, b]$ if: 1. $f$ is continuous at every interior point $c \in...
Full solved answer →Use continuity to evaluate the limit, $\lim_{x \to 4} \frac{5 + \sqrt{x}}{\sqrt{5 + x}}$ [5]
$$\lim{x \to 4} \frac{5 + \sqrt{x}}{\sqrt{5 + x}}$$ Point of evaluation: $a = 4$ --- Root functions, polynomial functions, and their sums, products, and quotients are continuous on their domains. If $f$ is continuous at $x = a$, then: $$\lim{x \to a} f(x) =...
Full solved answer →Precise definition of Limit
If a function is defined by $f(x) = \begin{cases} 1 + x, & x \leq -1 \ x^2, & x > -1 \end{cases}$, evaluate $f(-3)$, $f(-1)$ and $f(0)$ and sketch the graph. Prove that $\lim_{x \to 0} \frac{|x|}{x}$ does not exist. [10+0]
$$f(x) = \begin{cases} 1 + x, & x \leq -1 \\ x^2, & x -1 \end{cases}$$ Evaluate $f(-3)$, $f(-1)$, $f(0)$; sketch graph; prove $\lim{x\to 0}\frac{x}{x}$ does not exist. --- $f(-3)$: Since $-3 \leq -1$, use $1+x$: $$f(-3) = 1 + (-3) = -2$$ $f(-1)$: Since $-1 ...
Full solved answer →A function is defined by $f(x) = |x|$
Calculate $f(-3)$, $f(4)$, and sketch the graph.
Prove that the limit does not exist. $$\lim_{x \to 2} \frac{|x-2|}{x-2}$$ [5+0+5]
- Function: $f(x) = x$ - Required: $f(-3)$, $f(4)$, sketch of $y = x$ - Limit to analyze: $\displaystyle\lim{x \to 2} \frac{x-2}{x-2}$ - Marks split: [5 + 0 + 5] --- The absolute value function is defined piecewise as: $$f(x) = x = \begin{cases} x & \text{i...
Full solved answer →A function is defined by
$$f(x) = \begin{cases} x+2, & x<0 \ 1-x, & x>0 \end{cases}$$
Calculate $f(-1)$, $f(3)$, and sketch the graph. Prove that the limit does not exist.
$$\lim_{x \to 0} \frac{|x|}{x}$$
[5+0+5]
Given data: Piecewise function: $$f(x) = \begin{cases} x+2, & x<0 \\ 1-x, & x0 \end{cases}$$ Note: $f(0)$ is undefined. Tasks: - Compute $f(-1)$ and $f(3)$, sketch graph [5 marks] - Prove $\displaystyle\lim{x\to 0}\frac{x}{x}$ does not exist [5 marks] All d...
Full solved answer →Limits at infinity
Define limit of a function. $\lim_{x \to \infty} \left(x - \sqrt{x}\right)$ [5]
- Function to evaluate: $\lim{x \to \infty} \left(x - \sqrt{x}\right)$ - Also asked: definition of limit of a function. --- Let $f(x)$ be a function defined on an open interval containing the point $a$ (except possibly at $a$ itself). The limit of $f(x)$ as...
Full solved answer →Make Unit 2 stick
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