Limits

What is a Limit?

A limit describes the value that a function $f(x)$ approaches as the variable $x$ approaches a particular point $a$.

In formulas, a limit of a function is usually written as $\lim\limits_{x \to c} f(x) = L$, and is read as the limit of $f$ of $x$ as $x$ approaches $c$ equals $L$.

Let's consider the function $\mathrm{f}({x})={x}^2$

Observe that as $x$ takes values very close to $0$, the value of $f(x)$ is also close to $0$. (See graph below)

We can also interpret it in another way. If we input the values of $x$ which tend to approach $0$ (meaning close to $0$, either just smaller than $0$ or just larger than $0$ ), the value of $f(x)$ will tend to approach $0$(meaning close to $0$, either just smaller than $0$ or just larger than $0$).

Then we can say that, $\lim\limits_{x \to 0} f(x) = 0$

Similarly, when $x$ approaches $2$, the value of $f(x)$ approaches $4$, i.e. $\lim\limits_{x \to 2} f(x) = 4$ or $\lim\limits_{x \to 2} x^2 = 4$.

In general, as $x \to a$ , $f(x) \to l$, then $l$ is called the limit of the function $\mathrm{f}(\mathrm{x})$, which is symbolically written as $\lim\limits_{x \to a} f(x) = l$.

Now consider the function

$
f(x)=\frac{x^2-6 x-7}{x-7}
$
We can factor the function as shown
$f(x)=\frac{(x-7)(x+1)}{x-7} \quad$ [Cancel like factors in numerator and denominator.]

$
f(x)=x+1, x \neq 7
$

Notice that $x$ cannot be $7$, or we would be dividing by $ 0$ , so $7$ is not in the domain of the original function. To avoid changing the function when we simplify, we set the same condition, $x \neq 7$, for the simplified function. We can represent the function graphically

What happens at $x=7$ completely differs from what happens at points close to $x=7$ on either side. Just observe that as the input $x$ approaches $7$ from either the left or the right, the output approaches $8$. The output can get as close to $8$ as we like if the input is sufficiently near $7$ . So we say that the limit of this function at $x=7$ equals $8$.

So even if the function does not exist at x = a, still the limit can exist at that point as the limit is concerned only about the points close to $x=a$ and NOT at $x=a$ itself.

Some Properties of Limits:

• $\lim\limits_{\mathrm{x} \to \mathrm{a}} \mathrm{c} = \mathrm{c}$, where $c$ is a constant quantity.
• The value of $\lim\limits_{\mathrm{x} \to \mathrm{a}} {x} = \mathrm{a}$.
• Value of $\lim\limits_{x \to a} (b x + c) = b a + c$.
• $\lim\limits_{x \to a} x^n = a^n$, if $n$ is a positive integer.

Recommended Video Based on Limits

Solved Examples Based on Limits

Example 1: Which of the following is incorrect?

(1) As $x$ approaches $2$, tends to reach $4$.
2) As $x$ approaches tends to reach $0$.
3) As $x$ approach tends to reach $\infty$.

4) As $x$ approach $\frac{\pi}{2}$, then $\tan x$ has a tendency to reach $\infty$.

Solution:
Limits describe the behaviour of a function $f(x)$ as its variable $x$ approaches a particular number.

1) $\lim\limits_{x \to 2} x^2 = 4$ — Statement 1 is correct.

2) $\lim\limits_{x \to \pi} \sin x = 0$ — Statement 2 is correct.

3) $\lim\limits_{x \to \pi / 2} \sin x = 1$ — Statement 3 is incorrect.

4) $\lim\limits_{x \to \pi / 2} \tan x = \infty$ — Statement 4 is correct.
Hence, the answer is the option 3. $\qquad$

Example 2: $\lim _{x \rightarrow 0}\left(\left(\frac{\left(1-\cos ^2(3 x)\right.}{\cos ^3(4 x)}\right)\left(\frac{\sin ^3(4 x)}{\left(\log _e(2 x+1)^5\right.}\right)\right)$ is equal to [JEE MAINS 2023]
1) $24$
2) $9$
3) $18$
4) $15$

Solution:

$
\begin{aligned}
& \lim _{x \rightarrow 0}\left[\frac{1-\cos ^2 3 x}{9 x^2}\right] \frac{9 x^2}{\cos ^3 4 x} \cdot \frac{\left(\frac{\sin 4 x}{4 x}\right)^3 \times 64 x^3}{\left[\frac{\ln (1+2 x)}{2 x}\right]^5 \times 32 x^5} \\
& \lim _{x \rightarrow 0} 2\left(\frac{1}{2} \times \frac{9}{1} \times \frac{1 \times 64}{1 \times 32}\right)=18
\end{aligned}
$
Hence, the answer is the option (3).

Example 3: Let $a_1, a_2, a_3, \ldots, a_n n$ be $n$ positive consecutive terms of an arithmetic progression. If this is its common difference, then: $\lim\limits _{n \rightarrow \infty} \sqrt{\frac{d}{n}}\left(\frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\ldots \ldots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}\right)$ is

[JEE MAINS 2023]
1) $\frac{1}{\sqrt{d}}$
2) $1$
3) $\sqrt{d}$
4) $0$

Solution:

$
\begin{aligned}
& \lim\limits_{n \to \infty} \sqrt{\frac{d}{n}} \left( \frac{\sqrt{a_1} - \sqrt{a_2}}{a_1 - a_2} + \frac{\sqrt{a_2} - \sqrt{a_3}}{a_2 - a_3} + \cdots + \frac{\sqrt{a_{n-1}} - \sqrt{a_n}}{a_{n-1} - a_n} \right) \\
& = \lim\limits_{n \to \infty} \sqrt{\frac{d}{n}} \left( \frac{\sqrt{a_1} - \sqrt{a_2} + \sqrt{a_2} - \sqrt{a_3} + \cdots + \sqrt{a_{n-1}} - \sqrt{a_n}}{-d} \right) \\
& = \lim\limits_{n \to \infty} \sqrt{\frac{d}{n}} \left( \frac{\sqrt{a_n} - \sqrt{a_1}}{d} \right) \\
& = \lim\limits_{n \to \infty} \frac{1}{\sqrt{n}} \left( \frac{\sqrt{a_1 + (n - 1) d} - \sqrt{a_1}}{\sqrt{d}} \right) \\
& = \lim\limits_{n \to \infty} \frac{1}{\sqrt{d}} \left( \sqrt{\frac{a_1 + (n - 1) d}{n}} - \frac{\sqrt{a_1}}{n} \right) \\
& = 1
\end{aligned}
$

Hence, the answer is the option 2

Example 4: If $x$ approaches $2$, then the approximate value of is
1) $4$
2) $2$
3) $3$
4) $1$

Solution:

As we have learned

Condition on Limits -
The limit does not give actual value. It gives an approximate value.
- wherein
$f(x)=\frac{x^2+x-2}{x-1}$
$x$ is not defined at $\mathrm{x}=1$ but for $\mathrm{x}>1 \& \mathrm{x}<1$ it gives approximate values.

When x approaches $2, x-2$ simplifies to $\mathrm{x}+2$
Limit approaches to $4$
Hence, the answer is the option 1.

Example 5: If $x$ approaches $3$ , then $\frac{x^2-5 x+6}{x^2-4 x+3}$ has approximate value
1) $\frac{1}{2}$
2) $0$
3) $1$
4) $\frac{3}{2}$

Solution:

Condition on Limits -

The limit does not give actual value. It gives an approximate value.
- wherein

$
f(x)=\frac{x^2+x-2}{x-1}
$

$x$ is not defined at $x=1$ but for $x>1 \& x<1$ it gives approximate values.

$
\frac{x^2-5 x+6}{x^2-4 x+3}=\frac{(x-2)(x-3)}{(x-1)(x-3)}
$
When x approaches $3, \frac{x^2-5 x+6}{x^2-4 x+3}$ simplifies to $\frac{x-2}{x-1}$
$\therefore \frac{x-2}{x-1}$ approaches $\frac{3-2}{3-1}=1 / 2$
Hence, the answer is the option 1.