Introduction To Limits

I. Introduction:
Let us take a look at the function
$f(x) = \dfrac{x^2 - 4}{x - 2}$

As such, there is no difficulty in evaluating this function for all values of $x$, except for the value $x = 2$
For all other values of $x$, the function evaluates to:
$f(x) = \dfrac{(x - 2)(x + 2)}{x - 2}$.
Since $x \ne 2$, $x - 2 \ne 0$ and we can safely cancel it out from the numerator and the denominator and get:
$f(x) = x + 2$

Now let us see what happens at $x = 2$
Our ability to cancel out $(x - 2)$ mathematically remains correct even if $x$ deviates from $2$ by an infinitesimally small amount.
Let us call this infinitesimally small amount of change in $x$ as $\Delta x$
Therefore, our function becomes:
$f(2 + \Delta x) = \dfrac{\{(2 + \Delta x) + 2\}\{(2 + \Delta x) - 2\}}{(2 + \Delta x) - 2}$
$= \dfrac{\{4 + \Delta x\}\{\Delta x\}}{\Delta x}$

Since, $\Delta x$ is infinitesimally small but still not $0$, we should be able to cancel out $\Delta x$ in our fraction and be left with:
$f(2 + \Delta x) = 4 + \Delta x$ 

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Since $\Delta x$ is infinitesimally small, we can ignore it on both sides of this equation, and write:
$\displaystyle{\lim_{x \to 2}f(x)}$
(The above statement is read as $\unicode{0x201C} \text{Limit x tends to 2, f of x} \unicode{0x201D}$

$= \displaystyle{\lim_{x \to 2}\dfrac{(x + 2)(x - 2)}{x - 2}}$

$= \displaystyle{\lim_{x \to 2}\dfrac{(x + 2)(x - 2)}{x - 2}}$

$= \displaystyle{\lim_{\Delta x \to 0}\dfrac{(2 + \Delta x + 2)(2 + \Delta x - 2)}{2 + \Delta x - 2}}$

$= \displaystyle{\lim_{\Delta x \to 0}\dfrac{(4 + \Delta x)(\Delta x)}{\Delta x}}$

$= \displaystyle{\lim_{\Delta x \to 0}(4 + \Delta x)}$

$= 4$

II. Left And Right Side Limits:
Here we will take a look at the limit of a function $f(x)$ at the point $x = a$ as $x$ approaches $a$ from the left hand side as well as from the right hand side.
The notation commonly used here, is:
$\lim\limits_{x \rightarrow a^-}f(x)$ for left hand limit, which, in other words, means as $x$ approaches $a$ from the left hand (negative) side, and

$\lim\limits_{x \rightarrow a^+}f(x)$ for the right hand limit, which means as $x$ approaches $a$ from the right hand (positive) side.

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Formally, we can define the left hand limit $L$ as,
$\lim\limits_{x \rightarrow a^-}f(x) = L$ if for any $\epsilon \gt 0$, there exists $\delta(\epsilon)$ such that for all $0 \lt a - x \lt \delta(\epsilon)$ the inequality $|f(x) - L| \lt \epsilon$ holds true.

Similarly, for the right hand limit $B$ as,
$\lim\limits_{x \rightarrow a^+}f(x) = R$ if for any $\epsilon \gt 0$, there exists $\delta(\epsilon)$ such that for all $0 \lt x - a \lt \delta(\epsilon)$ the inequality $|f(x) - R| \lt \epsilon$ holds true

If $\lim\limits_{x \rightarrow a^-}f(x) = \lim\limits_{x \rightarrow a^+}f(x) = A$ for some finite value $A$, then we can say that:
$\lim\limits_{x \rightarrow a}f(x) = A$

If, on the other hand, $\lim\limits_{x \rightarrow a^-}f(x) \ne \lim\limits_{x \rightarrow a^+}f(x)$, but both limits exist, then $\lim\limits_{x \rightarrow a}f(x)$ is not defined.

The above statements means that for an assigned value $a$ of the independent variable $x$, if the value of the function can be brought arbitrarily close to a finite value $A$ of the function, then that value of the function is called the limit of the function as the variable $x$ approaches the value $a$.

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Now try this problem:
--------- Reference to question: cfdd67f4-e609-48ba-83a4-9a1260f349fc ---------
III. Properties Of Limits
If for two functions $f(x)$ and $g(x)$, both $\lim\limits_{x \rightarrow a} f(x)$ and $\lim\limits_{x \rightarrow a} g(x)$ exist, then the limit of their composite functions obey the following rules:
$\lim\limits_{x \rightarrow a} [f(x) \pm g(x)] = \lim\limits_{x \rightarrow a} f(x) \pm \lim\limits_{x \rightarrow a} g(x)$

$\lim\limits_{x \rightarrow a} [f(x) \cdot g(x)] = \lim\limits_{x \rightarrow a} f(x) \cdot \lim\limits_{x \rightarrow a} g(x)$

$\lim\limits_{x \rightarrow a} \dfrac{f(x)}{g(x)} = \dfrac{\lim\limits_{x \rightarrow a} f(x)}{\lim\limits_{x \rightarrow a} g(x)}$ if $\lim\limits_{x \rightarrow a} g(x) \ne 0$

Limits Of Composite Functions:
$\lim\limits_{x \rightarrow c} f(g(x)) = \lim\limits_{x \rightarrow L} f\left( \lim\limits_{x \rightarrow c} g(x)\right)$, where
$L = \lim\limits_{x \rightarrow c} g(x)$ and $L$ exists.

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Limits Of Exponent Functions:
$\lim\limits_{x \rightarrow a} f(x)^{g(x)}$ when $f(x)$ is continuous in $(0, +\infty)$

Let $L = \lim\limits_{x \rightarrow a} f(x)^{g(x)}$
$\Rightarrow \ln L = \ln \left( \lim\limits_{x \rightarrow a} f(x)^{g(x)} \right)$

$= \lim\limits_{x \rightarrow a} \left(\ln \left( f(x)^{g(x)} \right)\right)$

$= \lim\limits_{x \rightarrow a} \left(\ln \left( f(x)^{g(x)} \right)\right)$

$= \lim\limits_{x \rightarrow a} g(x) \cdot \left(\ln f(x)\right)$

$= \lim\limits_{x \rightarrow a} (g(x)) \cdot \left(\lim\limits_{x \rightarrow a} (\ln f(x))\right)$

$= \lim\limits_{x \rightarrow a} (g(x)) \cdot \ln \left(\lim\limits_{x \rightarrow a} f(x)\right)$

$=  \ln \left[\left(\lim\limits_{x \rightarrow a} f(x)\right)^{\lim\limits_{x \rightarrow a} (g(x))}\right]$

$=  \ln \left[\left(\lim\limits_{x \rightarrow a} f(x)\right)^{\lim\limits_{x \rightarrow a} (g(x))}\right]$

$\therefore L = \left(\lim\limits_{x \rightarrow a} f(x)\right)^{\lim\limits_{x \rightarrow a} (g(x))}$


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Now try the following problem:
--------- Reference to question: 2e30ad52-1d59-45e3-9a0f-bce0aa8ee0db ---------

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IV. Frequently Used Limits
The following are some of the limits that are frequently used to solve more complex problems related to limits.

$\left. \begin{array}{l} \lim\limits_{x \rightarrow 0} \dfrac{\sin x}{x} = 1 \\ \  \\ \lim\limits_{x \rightarrow 0} \dfrac{1 - \cos x}{x} = 1 \end{array} \right\}$   (both derivations can be found )

$\lim\limits_{x \rightarrow \infty} \left(1 + \dfrac{1}{x}\right)^x = e$   (the proof can be read .)

$\left. \begin{array}{l} \lim\limits_{x \rightarrow 0} \dfrac{\log_a(1 + x)}{x} = \log_a e \\ \ \\  \lim\limits_{x \rightarrow 0} \dfrac{\ln (1 + x)}{x} = 1 \\ \ \\ \lim\limits_{x \rightarrow 0} \dfrac{(a^x - 1)}{x} = \ln a \phantom{000} (a \gt 0) \end{array}\right\}$ (the proofs can be found .)

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Now, let us try the following problem:
--------- Reference to question: a98838cf-e849-4b6f-abf3-614086774885 ---------