CBSE Class 11 - Mathematics - Limits and Derivatives (Part-2) Questions and Answers #class11Maths #Limits #eduvictors

CBSE Class 11 - Mathematics - Limits and Derivatives (Part-2) 

Questions and Answers                                                                     #class11Maths #Limits 

In the previous blog post Limits and derivatives Part-1 , we learn

$\lim_{x\rightarrow a} f(x) = l$ and it is called limit of the function f(x)

① The two ways x could approach a number an either from left or from right, i.e., all the values of x near a could be less than a or could be greater than a.

② In this case the right and left hand limits are different, and hence we say that the limit of f(x) as x tends to zero does not exist (even though the function is defined at 0).

Find a limit when f(a) has definite value:

To find limit when f(a) has definite value, then $\lim_{x\rightarrow a} f(x) = f(a)$ i.e. direct substitution.

Q1: Find limits of the following:

⒜ $\lim_{x\rightarrow 0} sin{x}$

⒝ $\lim_{x\rightarrow-1} \frac{x^{10} + x^5 + 1}{x - 1}$

⒞ $\lim_{x\rightarrow 1}\left [ x^3 -x^2 + 1 \right ]$

⒟   $\lim_{x\rightarrow 1}\frac{ax^2 + bx + c}{cx^2 + bx + a}, a + b + c \neq 0$

Answer:

⒜ $\lim_{x\rightarrow 0} sin{x}$

= sin 0  = 0

⒝ $\lim_{x\rightarrow-1} \frac{x^{10} + x^5 + 1}{x - 1}$

= $\frac{(-1)^{10} + (-1)^5 + 1}{-1 - 1}$

= $\frac{1 -1 + 1}{-2}$ 

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

⒞ $\lim_{x\rightarrow 1}\left [ x^3 -x^2 + 1 \right ]$

=  $1^3 -1^2 + 1$ = 1

⒟  $\lim_{x\rightarrow 1}\frac{ax^2 + bx + c}{cx^2 + bx + a}, a + b + c \neq 0$

= $\frac{a(1)^2 + b(1) + c}{c(1)^2 + b(1) + a}$

= $\frac{a + b + c}{c + b + a}$

= $1 \because a + b + c \neq 0$

Find a limit when f(a) is inderminate (e.g. $\frac{0}{0} form$)

When f(a) is inderminate, wew follow different rules or approaches. The simplest one is factorize the numerator and the denominator. Cancel out the common factors and then put x = a.

Q2: Evaluate $\lim_{x \rightarrow 3} \left ( \frac{x^2 - 9}{x - 3} \right )$

Answer: 

$ \lim_{x \rightarrow 3} \left ( \frac{x^2 - 9}{x - 3} \right ) $

 

= $\lim_{x \rightarrow 3} \left ( \frac{(x+3)(x-3)}{x - 3} \right )$ 

= $\lim_{x \rightarrow 3} ( x+3 ) $

= 3 + 3 = 6

Before we go deeper in Algebra of limits, let us go through its theorems and properties:

① Theorem I. Uniqueness theorem

i. If $\lim_{x \rightarrow c^-} f(x) = l \text{ and } \lim_{x \rightarrow c^-} f(x) = l' \text{ then } l = l' $

ii. If  $\lim_{x \rightarrow c^+} f(x) = l \text{ and } \lim_{x \rightarrow c^+} f(x) = l' \text{ then } l = l' $

iii. If $\lim_{x \rightarrow c} f(x) = l \text{ and } \lim_{x \rightarrow c} f(x) = l' \text{ then } l = l' $

② Theorem II

$\lim_{x \rightarrow c} f(x)  \text{ exists iff } \lim_{x \rightarrow c^-}  \text{ and }  \lim_{x \rightarrow c^+} \text{ both exist and are equal}$

Properties of Limits

Assume that $\lim_{x \rightarrow a} f(x)$ and $\lim_{x \rightarrow a} g(x)$ exist. Let c be any real number, then the following are true:

① $\lim_{x \rightarrow a}\left [ f(x) \pm g(x) \right ] = \lim_{x \rightarrow a} f(x) \pm \lim_{x \rightarrow a} g(x)$

It means, the limit of a sum or difference is the sum or difference of limits.

② $\lim_{x \rightarrow a}\left [ f(x) g(x) \right ] = \lim_{x \rightarrow a} f(x) \times \lim_{x \rightarrow a} g(x)$

It means the limit of a product is the product of limits.

③ $\lim_{x \rightarrow a}\frac{f(x)}{g(x)} = \frac{\lim_{x \rightarrow a} f(x)}{\lim_{x \rightarrow a} g(x)} \text{ provided } \lim_{x \rightarrow a}g(x) \neq 0$

It says the limit of a quotient is the quotient of limits, provided the denominator does not have a limit 0.

④ $\lim_{x \rightarrow a}[cf(x)] = c. \lim_{x \rightarrow a}f(x)$

That is, the limit of a constant times a function is the constant times the limit.

⑤ If f(x) = c for all x, the $\lim_{x \rightarrow a}f(x) = c$.

It says, the limit of a constant function is a contant. Symbolically is it written as $\lim_{x \rightarrow a} c = c$

⑥ $\lim_{x \rightarrow a} \sqrt[n]{f(x)} =  \sqrt[n]{\lim_{x \rightarrow a}f(x)}$ for any positive integer n.

⑦ $\lim_{x \rightarrow a} x^n = a^n$ for any positive integer n.

In the next post, we shall solve some examples to evaluate algebraic limits.  

👉See Also:

Ch13 Limits and Derivatives - Q & A (Part-1)

SETS (NCERT Ex 1.1)
SETS (NCERT Ex 1.2)
Different Types of Sets
SETS (NCERT Ex 1.3)

SETS (Unit Test Paper)
SETS (VENN DIAGRAMS)
SETS (Operations of Sets)
SETS (NCERT Ex 1.4 Q1-Q5)
SETS (NCERT Ex 1.4 Q6 - Q8)
SETS (NCERT Ex 1.4 Q9 - Q12)
SETS (NCERT Ex 1.5)
SETS (NCERT Ex 1.6) 
Laws of Set Operations
Ch2: Relations and Functions (1 Mark Q & A) Part-1
Ch2: Cartesian Product of Two Sets (Important Points)
Ch2: Relations - Domain, Range and Co-Domain (Solved Problems)

Ch5: Complex Numbers (Part 1) - Solved Problems

Maths Annual Test Paper (2018-19)
Maths Annual Test Paper (2020-21) 
Maths Term1 MCQs (2021-22)