# CBSE Class 11 - Mathematics - Limits and Derivatives

Part-3  Algebra of Limits

In the previous post Limits and derivatives Part-2, we get basic ideas of the algebra of limits and also learned about rules and properties of limits.

Let us try to solve few problems:

Q1: Evaluate the given limit $\lim_{r\rightarrow 1} \pi r^2$

Answer: $\lim_{r\rightarrow 1} \pi r^2 = \pi(1)^2 = \pi$

Q2: Evaluate the given limit $\lim_{x\rightarrow 1} \frac{x^3-1}{x-1}$

$\lim_{x\rightarrow 1} \frac{x^3-1}{x-1}$

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

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

= $1^2 + 1 + 1 = 3$

Q3: Evaluate the following limit:

$\lim_{x\rightarrow 2} \frac{x^3-3x+2}{x^2 + x - 6}$

$\lim_{x\rightarrow 2} \frac{x^3-3x+2}{x^2 + x - 6}$

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

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

=$\frac{2-1}{2+3}$

=$\frac{1}{5}$

Let us prove another theorem

Theorem: For any positive integer n, (or nis any positive rational number)

$\lim_{x\rightarrow a} \frac{x^n-a^n}{x - a} =na^{n-1}$

Proof:

$\because x^n - a^n = (x - a)(x^{n-1} + x^{n-2}a + ... + xa^{n-2} + a^{n-1})$

$\lim_{x\rightarrow a} \frac{x^n-a^n}{x - a} =na^{n-1}$

= $lim_{x\rightarrow a} \frac{(x - a)(x^{n-1} + x^{n-2}a + ... + xa^{n-2} + a^{n-1})}{x-a}$

= $lim_{x\rightarrow a} (x^{n-1} + x^{n-2}a + ... + xa^{n-2} + a^{n-1})$

= $a^{n-1} + a^{n-2}.a + a^{n-3}.a^2+ ... + a.a^{n-2} + a^{n-1}$

= $a^{n-1} + a^{n-1} + a^{n-1} + ... + a^{n-1} + a^{n-1}$

= $n a^{n-1}$

Q4: Evaluate the given limit $\lim_{x\rightarrow 0}\frac{(x+1)^5 - 1}{x}$

Answer: Let x + 1 = y

Since x→0, therefore y→1

We have,

$\lim_{x\rightarrow 0}\frac{(x+1)^5 - 1}{x}$

$= \lim_{y\rightarrow 1}\frac{y^5 - 1}{y -1}$

Since, $\lim_{x\rightarrow a} \frac{x^n-a^n}{x - a} =na^{n-1}$

= $5.1^{5-1}$

= 5

Q5: Evaluate the given limit
$\lim_{z\rightarrow 1}\frac{z^\frac{1}{3} - 1}{z^\frac{1}{6} -1}$

Answer: At z = 1 the given function becomes indeterminate.
Let $z\frac{1}{6} = x$.
Since z→1, therefore x→1

= $\lim_{z\rightarrow 1} \frac{x^2 -1}{x - 1}$

Since, $\lim_{x\rightarrow a} \frac{x^n-a^n}{x - a} =na^{n-1}$

= $2.1^{2-1}$

= 2

Q6: Evaluate $\lim_{x\rightarrow 2} \frac{x^{5} - 32}{x^3 - 8}$

$\lim_{x\rightarrow 2} \frac{x^{5} - 32}{x^3 - 8}$
= $\lim_{x\rightarrow 2} \frac{x^{5} - 2^5}{x^3 - 2^3}$
= $\frac{\lim_{x\rightarrow 2}(x^{5} - 2^5)}{\lim_{x\rightarrow 2}(x^3 - 2^3)}$
= $\frac{\lim_{x\rightarrow 2}\frac{(x^{5} - 2^5)}{x-2}}{\lim_{x\rightarrow 2}\frac{(x^3 - 2^3)}{x-2}}$

Since, $\lim_{x\rightarrow a} \frac{x^n-a^n}{x - a} =na^{n-1}$

=$\frac{5\times2^4}{3 \times2^2}$
=$\frac{80}{12}$
=$\frac{20}{3}$

Q7: Evaluate $\lim_{x\rightarrow a}\frac{(x^{m} - a^m)}{(x^{n} - a^n)}$

$\lim_{x\rightarrow a}\frac{(x^{m} - a^m)}{(x^{n} - a^n)}$
= $\lim_{x\rightarrow a}\frac{(x^{m} - a^m)}{(x^{n} - a^n)}$
= $\frac{\lim_{x\rightarrow a}(x^m - a^m)}{\lim_{x\rightarrow a}(x^n - a^n)}$
= $\frac{\lim_{x\rightarrow a}\frac{(x^m - a^m)}{x-a}}{\lim_{x\rightarrow a}\frac{(x^n - a^n)}{x-a}}$

Since, $\lim_{x\rightarrow a} \frac{x^n-a^n}{x - a} =na^{n-1}$

=$\frac{m.a^{m-1}}{n. a^{n-1}}$
=$\frac{m}{n}a^{m-n}$

๐Q6 can be solved directly by using the above-said formula.

In the next post, we'll learn about trigonometry limits.