Limit of Algebraic function - Practice Questions & MCQ

Limit of Algebraic function

(b) Factorization Method

In this method, we factorize numerators and denominators. The common factors are canceled out and the rest of the output is the final answer. 

Illustration 1:

To evaluate $\lim _{x \rightarrow 2} \frac{x^3-2 x-4}{x^2-3 x+2}$
we can re-write,

$
\Rightarrow \lim _{x \rightarrow 2} \frac{(x-2)\left(x^2+2 x+2\right)}{(x-2)(x-1)} \quad\left[\frac{0}{0} \text { form }\right]
$

$(x-2)$ will cancel out in numerator and denominators

$
\Rightarrow \lim _{x \rightarrow 2} \frac{\left(x^2+2 x+2\right)}{(x-1)}=10
$

(c) Rationalization Method

This method is used when either numerator or denominator or both have fractional powers (like ½, ⅓ etc). After rationalization, the terms are factorized which on cancellation gives the final answer.

Let’s go through an illustration to understand better 

Illustration 2:

Evaluate : $\lim _{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}}$
Rationalizing numerator and denominator we get,

$
\begin{array}{ll}
=\lim _{x \rightarrow a} \frac{a+2 x-3 x}{3 a+x-4 x}\left(\frac{\sqrt{3 a+x}+2 \sqrt{x}}{\sqrt{a+2 x}+\sqrt{3 x}}\right) & \left(\frac{0}{0} \text { form }\right) \\
=\lim _{x \rightarrow a} \frac{a-x}{3(a-x)}\left(\frac{\sqrt{3 a+x}+2 \sqrt{x}}{\sqrt{a+2 x}+\sqrt{3 x}}\right) & \left(\frac{0}{0} \text { form }\right) \\
=\frac{1}{3}\left(\frac{2 \sqrt{a}+2 \sqrt{a}}{\sqrt{3 a}+\sqrt{3 a}}\right) & \\
=\frac{2}{3 \sqrt{3}} &
\end{array}
$
 

Limit of Algebraic Function Using Standard Result

As we studied in the Binomial Theorem, the binomial expansion for any index

$
(1+x)^n=1+n x+\frac{n(n-1)}{2!} x^2+\frac{n(n-1)(n-2)}{3!} x^3 \ldots
$

where, $|x|<1$
When $x$ is infinitely small (approaching zero), we can ignore higher powers of $x$ and can write $(1+x)^n$ $=1+n x$ (approximately).

We will use this fact to evaluate the value of

$
\lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}
$
Let $\mathrm{x}=\mathrm{a}+\mathrm{h}$,

$
\begin{aligned}
\therefore \lim _{x \rightarrow a} \frac{x^n-a^n}{x-a} & =\lim _{h \rightarrow 0} \frac{(a+h)^n-a^n}{(a+h)-a} \\
& =\lim _{h \rightarrow 0} \frac{a^n\left\{\left(1+\frac{h}{a}\right)^n-1\right\}}{h} \\
& =a^n \lim _{h \rightarrow 0} \frac{\left\{1+n \frac{h}{a}\right\}-1}{h}
\end{aligned}
$

$\left[\right.$ when $\left.\mathrm{x} \rightarrow 0,(1+x)^n \rightarrow 1+n x\right]$

$
\begin{aligned}
& =a^n \cdot \frac{n}{a}=n a^{n-1} \\
\therefore \lim _{\mathbf{x} \rightarrow \mathbf{a}} \frac{\mathbf{x}^{\mathbf{n}}-\mathbf{a}^{\mathbf{n}}}{\mathbf{x}-\mathbf{a}} & =\mathbf{n a}^{\mathbf{n}-\mathbf{1}}
\end{aligned}
$
Also, $\quad \lim _{x \rightarrow 0} \frac{(1+x)^n-1}{x}=n$

Illustration 1: 

If $\lim _{x \rightarrow 3} \frac{x^n-3^n}{x-3}=108$, find the value of $n$ such that $n \in \mathbb{N}$
Solution:
we have $\lim _{\mathrm{x} \rightarrow 3} \frac{\mathrm{x}^{\mathrm{n}}-3^{\mathrm{n}}}{\mathrm{x}-3}=108$,

$
\begin{aligned}
& \Rightarrow \mathrm{n} \cdot 3^{\mathrm{n}-1}=108=4 \times 3^3 \\
& \Rightarrow \mathrm{n}=4
\end{aligned}
$

Algebraic Function of type ‘infinity/infinity’

To find the limit of the type $\mathrm{x} \rightarrow \infty$, write the given expression in the form of $\frac{N}{D}(\mathrm{D} \neq 0),(\mathrm{N}$ is the numerator, D is the denominator). Then divide both N and D by the highest power of x occurring in both N and D to get a meaningful form.
An important result:
If $m, n$ are positive integers and $a_0, b_0 \neq 0$ are non-zero real numbers, then

$
\lim _{x \rightarrow \infty} \frac{a_0 x^m+a_1 x^{m-1}+\ldots+a_{m-1} x+a_m}{b_0 x^n+b_1 x^{n-1}+\ldots+b_{n-1} x+b_n}= \begin{cases}0 & \text { if } m<n \\ \frac{a_0}{b_0} & \text { if } m=n \\ \infty & \text { if } m>n \quad \text { when } a_0 b_0>0 \\ -\infty & \text { if } m>n \text { when } a_0 b_0<0\end{cases}
$