Question

Does the limit exist? If the limit does exists what is the value?Calculus early transcendental functions

Answer 1

To determine if a limit exists, you can evaluate the function at that point or use algebraic methods. If the function approaches a specific value as the input approaches a particular point, then the limit exists and the value of the limit is that specific value.

Step-by-step explanation:

The question asks whether a limit exists and if it does, what its value is. In calculus, limits are used to describe the behavior of functions as they approach a certain point or infinity. To determine if a limit exists, you can evaluate the function at that point or use algebraic methods such as factoring or rationalization. If the function approaches a specific value as the input approaches a particular point, then the limit exists and the value of the limit is that specific value.

For example, if we consider the function f(x) = x^2, we can see that as x approaches 2, the function approaches 4. Therefore, the limit of f(x) as x approaches 2 is 4. On the other hand, if we consider the function g(x) = 1/x, as x approaches 0, the function approaches infinity. So, the limit of g(x) as x approaches 0 is infinity.

Answer 2

given

`\lim _(x\to\infty)(x^2+2)/(1-x-3x^2)`
`\begin{gathered} \lim _(x\to\infty)(x^2+2)/(1-x-3x^2) \\ \lim _(x\to\infty)(1+(2)/(x^2))/((1)/(x^2)-(x)/(x^2)-3) \\ \lim _(x\to\infty)(1+(2)/(x^2))/((1)/(x^2)-(1)/(x)-3) \end{gathered}`

let x tends to infinty

`\begin{gathered} \lim _(x\to\infty)(1+(2)/(x^2))/((1)/(x^2)-(1)/(x)-3) \\ (1+(2)/(\infty^2))/((1)/(\infty^2)-(1)/(\infty)-3) \end{gathered}`

since ,1 divided by infinty is underfined.

limit does not exist.