1.3. Limits and [[Continuity]]

The equation of a limit can be written as lim𝑥→𝑎𝑓(𝑥)=𝐴lim𝑥→𝑎𝑓(𝑥)=𝐴.

It indicates that as 𝑥𝑥 approaches ’𝑎𝑎_’_ both from the left of 𝑥=𝑎𝑥=𝑎 and the right of 𝑥=𝑎𝑥=𝑎, the output value of the function approaches 𝐴𝐴.

For example, when the limit of 𝑥𝑥 is 2 for a function 𝑓(𝑥)=2𝑥+3𝑓(𝑥)=2𝑥+3, then the function will approach 𝐴=7𝐴=7.

We can write it as $lim𝑥→2(2𝑥+3)=7$.

What this says is that when the input values approach 2, the output values will get close to 7, i.e. as 𝑥 approaches 2 (but not exactly 2), the output function 𝑓(𝑥) gets a value closer to 7. Let’s explore this further.

Limits and continuity (2023) adapted from Faculty of Information Technology

Left-hand limits and Right-hand limits

Look at a function $𝑓(𝑥)=𝑥+3,𝑥≠5$.

Here, it deliberately indicates that 𝑥 cannot be assigned the value of 5. If we analyse the behaviour of 𝑓(𝑥):

They are indicated by $lim_{𝑥→5-}𝑓(𝑥)=8$ and $lim_{𝑥→5+}𝑓(𝑥)=8$

Limits can be found using graphs, too.

Calculating Limits

To calculate the limit of a function as $x$ approaches a certain value, follow these steps:

1. Direct Substitution: Substitute the value of $x$ into the function. If the result is a finite number, that is the limit. For example, $\lim_{x \to 2} (3x + 1) = 3(2) + 1 = 7$.

2. Simplify the Expression: If direct substitution results in an indeterminate form like $\frac{0}{0}$ or $\infty - \infty$, try to simplify the function. This might involve factoring, rationalising, or using algebraic identities.

3. Use [[L’Hôpital’s Rule]]: If simplification doesn’t work, and you still get an indeterminate form of the type $\frac{0}{0}$ or $\frac{\infty}{\infty}$, apply L’Hôpital’s Rule, which states that $\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f’(x)}{g’(x)}$, provided the latter limit exists.

4. Analyse One-Sided Limits: For limits as $x$ approaches a value from the left ($x \to a^-$) or the right ($x \to a^+$), evaluate the function by approaching the value from the respective side. Both one-sided limits must be equal for the two-sided limit to exist.

5. Check for Infinite Limits: If the function grows without bound as $x$ approaches a value, the limit might be $\infty$ or $-\infty$. Analyze the behaviour of the function to determine this.

6. Utilise Squeeze Theorem: If the function can be “squeezed” between two other functions that have the same limit at a point, use the Squeeze Theorem to find the limit.

Problem:

Evaluate the following:

$$lim_{𝑥→−∞}\frac{\sqrt{4𝑥^6+5}}{𝑥3+1}$$

$$lim_{𝑥→−5}\frac{𝑥^2−25}{𝑥^2+2𝑥−15}$$ by direct substitution x=-5 $$\frac{25-25}{25-10-15} = \frac{0}{0}$$ this is an indeterminate form so not valid by Factoring Quadratic Equations $$\text{discriminant}=b^2-4ac = 2^2-(4*-15) = 64 = 8^2$$

$$𝑥^2+2𝑥−15, ac=-15: m = 5, n= -3$$ $$x^2 + 5x - 3x - 15 = x(x+5) - 3(x+5) = (x-3)(x+5)$$

$$\frac{(x+5)(x-5)}{(x-3)(x+5)} = \frac{x-5}{x-3}$$ by direct substitution again x=-5 $$ \frac{-10}{-8} = \frac{5}{4}$$ $$lim_{𝑥→5}\frac{𝑥^3−125}{𝑥−5}$$ direct substitution is indeterminate $$lim_{𝑥→5}\frac{(x-5)(x^2 + 5x + 25)}{x-5}$$ $$lim_{𝑥→5}x^2 + 5x + 25$$ substitute x=5 $$25 + 25 + 25 = 75$$

$$lim_{𝑥→−4}(10𝑥+36)^3$$ by direct sub $$lim_{𝑥→−4}(-4)^3 = -64$$

$$lim_{𝑥→0}\frac{5sin(𝑥)}{3𝑥}$$ trying [[L’Hôpital’s Rule]]! $$lim_{𝑥→0}\frac{5cos(𝑥)}{3} = \frac{5}{3}$$