Given that limx →2 f(x)=4 limx →2 g(x)=-2 limx →2h(x)=0 find the limits that exist.

step 1 So we have the limit as x approaches to f of x equals 4. The limit as x approaches 2, G of x equ

Given that limx →2 f(x)=4 limx →2 g(x)=-2 limx...

Given that $$\lim _{x \rightarrow 2} f(x)=4 \quad\quad\quad \lim _{x \rightarrow 2} g(x)=-2 \quad\quad\quad \lim _{x \rightarrow 2}h(x)=0$$ find the limits that exist. If the limit does not exist, explain why. \begin{equation}\text { (a) }\lim _{x \rightarrow 2}[f(x)+5 g(x)] \quad \text { (b) } \lim _{x \rightarrow 2}[g(x)]^{3}\end{equation} \begin{equation} \begin{array}{ll}{\text { (c) } \lim _{x \rightarrow 2} \sqrt{f(x)}} & {\text { (d) } \lim _{x \rightarrow 2} \frac{3 f(x)}{g(x)}} \\ {\text { (e) } \lim _{x \rightarrow 2} \frac{g(x)}{h(x)}} & {\text { (f) } \lim _{x \rightarrow 2} \frac{g(x) h(x)}{f(x)}}\end{array} \end{equation}

Given that
$$\lim _{x \rightarrow 2} f(x)=4 \quad\quad\quad \lim _{x \rightarrow 2} g(x)=-2 \quad\quad\quad \lim _{x \rightarrow 2}h(x)=0$$
find the limits that exist. If the limit does not exist, explain
why.
\begin{equation}\text { (a) }\lim _{x \rightarrow 2}[f(x)+5 g(x)] \quad \text { (b) } \lim _{x \rightarrow 2}[g(x)]^{3}\end{equation}
\begin{equation}
\begin{array}{ll}{\text { (c) } \lim _{x \rightarrow 2} \sqrt{f(x)}} & {\text { (d) } \lim _{x \rightarrow 2} \frac{3 f(x)}{g(x)}} \\ {\text { (e) } \lim _{x \rightarrow 2} \frac{g(x)}{h(x)}} & {\text { (f) } \lim _{x \rightarrow 2} \frac{g(x) h(x)}{f(x)}}\end{array}
\end{equation}

Key Concepts

Indeterminate Forms and Undefined Limits

When evaluating limits, particularly those involving fractions, it is important to determine whether the denominator approaches zero. If it does, the expression might result in an indeterminate form or be undefined, requiring further analysis or different techniques to resolve the limit.

Algebraic Manipulation of Limits

The use of algebraic manipulation in limit problems involves applying properties such as distribution, factoring, and combining like terms to simplify limit expressions. This method relies on standard limit laws and enables the conversion of complex expressions into simpler forms where the limit can be directly computed.

Limit Laws

Limit laws provide a framework for computing the limits of functions by dividing complex expressions into simpler parts. These laws, including the sum, constant multiple, product, quotient, and power rules, are fundamental in calculus and allow one to combine the limits of individual functions to find the limit of an overall expression.

Continuous Functions

Continuous functions are those for which the limit as x approaches a certain point equals the function's value at that point. This concept is crucial when applying operations, like compositions or radicals, since if a function is continuous at a given value, the limit of the composite function can be directly obtained by substituting the limit of the inner function.

Transcript

00:03 So we have the limit as x approaches to f of x equals 4.

00:09 The limit as x approaches 2, g of x equals negative 2, the limit as x approaches 2, h of x equals 0.

00:22 So if we are giving these expression down here, we are to determine if the limit exists and if it does what is the value.

00:32 So let's get started.

00:35 So first we have a.

00:37 It says that the limit as x approaches to, what is the value of f of x plus five times of g of x.

00:47 So since it's an addition of two limits, we can separate the limits into two different parts.

00:55 So the limit as x approaches two of f of x plus the limit of the limit of the limit of x approaches 2 of 5 times g of x and since 5 is just a constant what we can do is we can move the 5 in front of the limit so all we have to do in the end is just multiplying 5 with whatever the limit is so we know that the limit as x approaches 2 f of x reaches 4 so this part right here equals 4 and the limit as x approaches 2, g of x will become negative 2.

01:37 So that will be negative 2.

01:39 And when we multiplied by 5, we're going to get negative 10.

01:44 So negative 10, which means that this equals negative 6.

01:49 So in the end, this limit exists and it is negative 6.

01:58 Let's move on to b.

01:59 So b as for the limit as x approaches to of g of x cubed.

02:11 What we can do here in this case is we move the limit operator into next to the function and then whatever we have there we're going to take the cube of it.

02:25 So we know that the limit as x approaches to g of x equals negative 2.

02:30 So this inside here will be negative 2 and when we cube it we're going to get negative 8...

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