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# Limits in calculus

Top
 Sub Topics Limits are the foundation of calculus. Before getting started with calculus, one must have the knowledge of limits. Limit of a function defines its nature, when the function approaches to a particular point. More specifically, limit of a function or a sequence is the value of that function or sequence, when it gets closer to a specified point. Limit of a function (or a sequence) is NOT its value at a definite point, rather than, it is the value of a function when it approaches nearest to a definite point. There are many concepts, such as differentiation, integration, continuity etc., which are described by the use of limits.

## Definition of a Limit

Formal definition of limits is known as $\epsilon$ (epsilon), $\delta$ (delta) definition of the limits which states that:
Let f be a function defined over an open interval which contains a value "a" in it.
Then, f(x) equals L when x tends to "a" or we write
$\lim_{x \rightarrow a}f(x) = L$

If for every real number $\epsilon > 0$, there exists a real number $\delta > 0$,
such that $\left | f(x) - L \right | < \epsilon$,
for all $0 < \left | x - a \right | < \delta$

Explanation:
Let us assume a function f(x) which is shown in the following diagram:

Definition of limits says that the value to f(x) is L, when x is nearest to a. This means, when value of x lies within the interval $(a - \delta, a + \delta)$ (where, $\delta$ is a real and positive number), eventually the value of f(x) lies within the interval $(L - \epsilon, L + \epsilon)$ (where, $\epsilon$ is a real and positive number).

## Limit of a Function

Limit of a function is known as its output value, when input approaches to a particular point. Let us consider a function f(x) which is defined upon an open interval, then limit of f(x) becomes L, when x tends to "a" which is contained in given interval. We can write that:
$\lim_{x \rightarrow a}f(x) = L$
This is pronounced as limit of f(x) as x approaches to (or tends to) a is L.
Let us consider an easy example:

### Solved Example

Question: Calculate $\lim_{x \rightarrow 2}(x^{2} + 1)$.
Solution:
Substituting value of x = 2 in the given function,
$\lim_{x \rightarrow 2}(x^{2} + 1)$
$= 2^{2} + 1 = 5$

## Limit of a Sequence

Limit of a sequence defines convergence and divergence of a sequence. Let us consider a sequence ${x_{n}}$ over the set of real numbers, then ${x_{n}}$ converges to L, when n approaches to a sufficiently large number. In other words,
$\lim_{n \rightarrow \infty }{x_{n}} = L$
If for every real number $\epsilon > 0$, there exists a natural number n $(n \geq N)$, such that $\left | x_{n} - L \right | < \epsilon$.
• If a sequence gives a finite value as it approaches to infinity, then the sequence is convergent or said to converge to that finite value.
• If a sequence gives an infinite value as it approaches to infinity, then the sequence is divergent or said to diverge.
Let us consider an easy example:

### Solved Example

Question: Determine if the sequence ${\frac{1}{n}}$, where $n\ \epsilon\ N$, converges or diverges.
Solution:
Sequence ${\frac{1}{n}}$ = 1, $\frac{1}{2}$, $\frac{1}{3}$, $\frac{1}{4}$, .....
$\lim_{n \rightarrow \infty } $$\frac{1}{n} = \frac{1}{\infty} = 0 The given sequence converges to zero. ## Limit Problems Back to Top Few problems based on limits are as follows: ### Solved Examples Question 1: Calculate \lim_{x \rightarrow 1}$$\frac{x(4x^{2} - 9)}{2x - 3}$.
Solution:
$\lim_{x \rightarrow 1}$$\frac{x(4x^{2} - 9)}{2x - 3} = \lim_{x \rightarrow 1}$$\frac{x((2x)^{2} - 3^{2})}{2x - 3}$

= $\lim_{x \rightarrow 1}$$\frac{x((2x)^{2} - 3^{2})}{2x - 3} = \lim_{x \rightarrow 1}$$\frac{x(2x + 3)(2x - 3)}{2x - 3}$

= $\lim_{x \rightarrow 1} \times (2x + 3)$
= $1 \times (2 \times 1 + 3)$
= 5

Question 2: A function f is defined as:
$f(x)=\left\{\begin{matrix} x^{2} &if\ x\geqslant 3 \\ 2x^{3}-7 &if\ x< 3 \end{matrix}\right.$
Find the value of $\lim_{x\rightarrow 3}f(x)$ and $\lim_{x\rightarrow 2}f(x)$.
Solution:
$\lim_{x\rightarrow 3}f(x)$ = $\lim_{x\rightarrow 3}x^{2}$
= $3^{2}$
= 9
$\lim_{x\rightarrow 2}f(x)$ = $\lim_{x\rightarrow 2}(2x^{3} - 7)$
= $2(2)^{3} - 7$
= 16 - 7
= 9

Question 3: Determine whether the following sequence is convergent:
$\frac{1}{3}$, $\frac{1}{5}$$\frac{1}{7}$$\frac{1}{9}$, .....
Solution:
Given sequence is
$\frac{1}{3}$$\frac{1}{5}$$\frac{1}{7}$$\frac{1}{9}, ..... nth term of the sequence would be: \left \{ x_{n} \right \} = \frac{1}{2n + 1} \lim_{n \to \infty }\left \{ x_{n} \right \} = \lim_{n \to \infty }$$\frac{1}{2n + 1}$
= $\frac{1}{\infty}$ = 0
Therefore, the given sequence is convergent.