Limits(part 2)

 Limits but x approaches to $\infty$ or -$ \infty$

What is Infinity?

Infinity is something which has no end or limit. It is not a number in the traditional sense, but a concept of never-ending.
In mathematics: Infinity ($\infty$) is used to describe quantities greater than any finite number. For example, the set of natural numbers (1, 2, 3,.) contains an infinite number.

If we consider $\lim_{x \to \infty}$ f(x) or $\lim_{x \to -\infty}$ f(x), we're considering the behavior of the function f(x) as the input x grows without bound in the positive or negative direction, respectively. Instead of x getting arbitrarily close to a number, it's getting arbitrarily large (positive or negative).
Consider the function f(x) = $\frac{1}{x}$.
As x becomes very large (for instance, x = 100, 1000, 10000), the value of $\\frac{1}{x}$ becomes very small (0.01, 0.001, 0.0001), near zero. We write this as:
$\lim_{x \to \infty}$ $\frac{1}{x}$ = 0
Similarly, as x becomes very large in the negative direction (for instance, x = -100, -1000, -10000), the value of $\\frac{1}{x}$ also approaches zero, but via negative values:
$\lim_{x \to -\infty}$ $\frac{1}{x}$ = 0
Consider another example:
Consider the function f(x) = $\frac{1}{x+1}$.
We want to find the limit as x approaches infinity:
$\lim_{x \to \infty}$ $\frac{1}{x+1}$
As x gets larger and larger (like 10, 100, 1000), the denominator $x+1$ gets larger and larger. So we have a fraction with the top constant at 1 and the bottom gigantic.
What's happening to the value of the fraction? It's getting closer and closer to zero.
For example:
If x = 10, $\frac{1}{10+1}$ = $\frac{1}{11}$ $\approx 0.09$
If x = 100, $\frac{1}{100+1}$ = $\frac{1}{101}$ $\approx 0.0099$
If x = 1000, $\frac{1}{1000+1}$ = $\frac{1}{1001}$ $\approx 0.000999$
As you can see, the value is getting closer to 0.
Therefore,
$\lim_{x \to \infty}$ $\frac{1}{x+1}$ = 0
Similarly, let's consider the limit as x approaches negative infinity:
$\lim_{x \to -\infty}$ $\frac{1}{x+1}$
Again if x takes a very large negative number (like -10, -100, -1000), the denominator $x+1$ also takes a very large negative number.
For example:
If x = -10, $\frac{1}{-10+1}$ = $\frac{1}{-9}$ $\approx -0.11$
If x = -100, $\frac{1}{-100+1}$ = $\frac{1}{-99}$ $\approx -0.0101$
If x = -1000, $\frac{1}{-1000+1}$ = $\frac{1}{-999}$ $\approx -0.001001$
The value is coming closer and closer to 0.
Therefore,
$\lim_{x \to -\infty}$ $\frac{1}{x+1}$ = 0
Then when limit comes infinity?
Let's think about the function f(x) = $x^3$.
We want to determine the limit as x approaches infinity:
$\lim_{x \to \infty} x^3$
As x gets larger and larger (such as 2, 10, 100), $x^3$ gets bigger even more quickly:
If x = 2, $x^3$ = $2^3$ = 8
If x = 10, $x^3$ = $10^3$ = 1000
If x = 100, $x^3$ = $100^3$ = 1,000,000
As x keeps going up and up without stopping, x^3 also goes up without stopping.
So,
$\lim_{x \to \infty}$ $x^3 = \infty$
Then let's consider the limit as x approaches negative infinity:
$\lim_{x \to -\infty} x^3$
As x becomes a very large negative number (like -2, -10, -100):
If x = -2, $x^3$ = $(-2)^3$ = -8
If x = -10, $x^3$ = $(-10)^3$ = -1000
If x = -100, $x^3$ = $(-100)^3$ = -1,000,000
As x gets more and more negative, $x^3$ gets more and more negative, without bound.
So,
$\lim_{x \to -\infty}$ $x^3 = -\infty$
Limit's intermediate form(eg. $\frac{0}{0}$ or $\frac{\infty}{\infty}$,etc)
If we substitute directly the value that x is approaching into a function, we sometimes obtain expressions which don't immediately inform us of the limit's value. These are called indeterminate forms. They tell us that we need to try harder (usually algebraic manipulation or the application of rules such as L'Hôpital's Rule) to find the actual limit.
The following are the indeterminate forms:
$\frac{0}{0}$: When $\lim_{x \to a}$ f(x) = 0 and $\lim_{x \to a}$ g(x) = 0, then $\lim_{x \to a}$ $\frac{f(x)}{g(x)}$ has the form $\frac{0}{0}$, an indeterminate form. We cannot say the limit is 0, $\infty$, or any other finite number.
Example: $\lim_{x \to 1}$ $\frac{x^2 - 1}{x - 1}$. Direct substitution gives $\frac{1^2 - 1}{1 - 1}$ = $\frac{0}{0}$. We can factor the numerator: $\lim_{x \to 1}$ $\frac{(x - 1)(x + 1)}{x - 1}$ = $\lim_{x \to 1}$ $(x + 1) = 2$.
$\frac{\infty}{\infty}$: If $\lim_{x \to a}$ f(x) = $\pm \infty$ and $\lim_{x \to a}$ g(x) = $\pm \infty$, then $\lim_{x \to a}$ $\frac{f(x)}{g(x)}$ is of the form $\frac{\infty}{\infty}$, which is also indeterminate.
Example: $\lim_{x \to \infty}$ $\frac{3x^2 + 1}{2x^2 - 5}$. Direct substitution gives $\frac{\infty}{\infty}$. We can divide numerator and denominator by $x^2$: $\lim_{x \to \infty}$ $\frac{3 + \frac{1}{x^2}}{2 - \frac{5}{x^2}}$ = $\frac{3 + 0}{2 - 0}$ = $\frac{3}{2}$.
$0 \cdot \infty$: If $\lim_{x \to a}$ f(x) = 0 and $\lim_{x \to a}$ g(x) = $\pm \infty$, then $\lim_{x \to a}$ f(x)g(x) is of the form $0 \cdot \infty$, which is indeterminate. We typically rewrite this as $\frac{0}{1/\infty} = \frac{0}{0}$ or $\frac{\infty}{1/0} = \frac{\infty}{\infty}$ in order to use methods for those cases.
Example: $\lim_{x \to 0^+}$ $x \ln x$. In this, $\lim_{x \to 0^+}$ x = 0 and $\lim_{x \to 0^+}$ $\ln x = -\infty$. We can write this in the form $\lim_{x \to 0^+}$ $\frac{\ln x}{1/x}$, which is of the form $\frac{-\infty}{\infty}$. L'Hôpital's Rule  would be useful in such a case.
$\infty - \infty$: When $\lim_{x \to a}$ f(x) = $\pm \infty$ and $\lim_{x \to a}$ g(x) = $\pm \infty$, then $\lim_{x \to a}$ $(f(x) - g(x))$ takes the form $\infty - \infty$, which is indeterminate. We typically need to combine the terms or use algebraic manipulation.
Example: $\lim_{x \to 0^+}$ $(\frac{1}{x} - \frac{1}{\sin x})$. Both terms approach $\infty$ as $x \to 0^+$. Their sum: $\lim_{x \to 0^+}$ $\frac{\sin x - x}{x \sin x}$, of the form $\frac{0}{0}$.

Why don't we keep the infinity as a output?

The point of importance is that "infinity" ($\infty$) is not a number; it's a concept for something having no bound. Because it is not an actual number, we cannot actually "keep" it as the ultimate value of an expression in normal analysis.
When we say $\lim_{x \to \infty}$ f(x) = L (where L is a real number), what we're saying is that as x gets arbitrarily large, f(x) gets arbitrarily close to the specific number L. We're talking about the direction the output of the function goes.
So in the same way, when we write $\lim_{x \to \infty}$ $f(x) = \infty$, we are stating that x gets arbitrarily large and the value of f(x) also goes without bound. We are stating that the value of the function also approaches infinity.
Thus, instead of "keeping infinity," we're trying to determine what becomes of the function as its input becomes indefinitely large. Does it approach a finite value? Does it likewise become indefinitely large? Or something else (such as oscillating)?
Finding the "value of the limit as x approaches infinity" is actually discovering this behavior. We want to know if and where the function settles down to a particular number or if it continues on (or decreases) forever.
Suppose you are watching a car move into the distance. We don't say where its eventual destination is "infinity." Instead, we might talk of its speed or direction of movement. If its speed is approaching some constant value, that's a limit. If it keeps speeding up, that's a limit of infinity.

L'Hospitals Rule:

L'Hôpital's Rule is a handy calculus method for finding the limits that yield indeterminate forms. Such forms typically arise when direct substitution yields expressions like $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

By the rule, if we have a limit of the form $\lim_{x \to c}$ $\frac{f(x)}{g(x)}$ where either:

$\lim_{x \to c}$ f(x) = 0 and $\lim_{x \to c}$ g(x) = 0, or

$\lim_{x \to c}$ f(x) = $\pm \infty$ and $\lim_{x \to c}$ g(x) = $\pm \infty$,

and if the limit $\lim_{x \to c}$ $\frac{f'(x)}{g'(x)}$ exists, then:

$\lim_{x \to c}$ $\frac{f(x)}{g(x)}$ = $\lim_{x \to c}$ $\frac{f'(x)}{g'(x)}$

That is, if you get an indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, you can differentiate the numerator and the denominator separately and then reevaluate the limit.

Example for $\frac{0}{0}$:

Consider $\lim_{x \to 1}$ $\frac{x^2 - 1}{x - 1}$. Direct substitution gives $\frac{1^2 - 1}{1 - 1}$ = $\frac{0}{0}$. Applying L'Hôpital's Rule:

$\lim_{x \to 1} \frac{\frac{d}{dx}(x^2 - 1)} {\frac{d}{dx}(x - 1)}$ = $\lim_{x \to 1} \frac{2x}{1}$ = $2(1) = 2$

Example for $\frac{\infty}{\infty}$:

Consider $\lim_{x \to \infty}$ $\frac{e^x}{x}$. Direct substitution gives \frac{\infty}{\infty}. Applying L'Hôpital's Rule:

$\lim_{x \to \infty} \frac{\frac{d}{dx}(e^x)} {\frac{d}{dx}(x)}$ = $\lim_{x \to \infty}$ $\frac{e^x}{1}$ = $\infty$

You can reapply L'Hôpital's Rule if the subsequent limit is still indeterminate.

Keep in mind that you are only taking the derivative of the numerator and denominator separately, and not the quotient rule! In addition, the rule only works explicitly with the indeterminate forms $\frac{0}{0}$ and $\frac{\infty}{\infty}$. All the other indeterminate forms such as $0 \cdot \infty$, $\infty - \infty$, $1^\infty$, $0^0$, and $\infty^0$ must first be algebraically rearranged to one of these two forms prior to the application of L'Hôpital's Rule. For example, $0 \cdot \infty$ can sometimes be written as $\frac{0}{1/\infty}$ = $\frac{0}{0}$ or as $\frac{\infty}{1/0}$ = $\frac{\infty}{\infty}$. Exponential forms are often solved with logarithms used to reduce the exponent.

L'Hôpital's Rule provides a systematic solution to limits that are otherwise problematic or require more advanced algebraic techniques.

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