Wednesday, 10 November 2021

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

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

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


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)


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.


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 substituition.


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:


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

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


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

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


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

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


$\lim_{x \rightarrow a}[cf(x)] = c. \lim_{x \rightarrow a}f(x)$
That is, the limit of a constant times a function is the constant times the limit.

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 $\lim_{x \rightarrow a} c = c$

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

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


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


👉See Also:

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