Learn rules of different functions of direct substitution, limits of absolute value function, rationalize the numerator, Evaluating left and right hand limits and greatest integer function.

- If 'g' is a polynomial or a rational function, then \(\lim\limits_{x\rightarrow a}g(x)=g(a)\)(directly put the value of 'a' in the function).
- This assumes that x = a is in the domain of 'g'.

A \(\dfrac {23}{8}\)

B 17

C \(\dfrac {-35}{9}\)

D \(\dfrac {-23}{11}\)

Consider,

\(\lim\limits_{x\rightarrow 2}\;\dfrac {x-2}{x^2-5x+6}\)

- If we put x = 2, we get \(\dfrac {0}{0}\) which is meaningless, so x = 2 is not in the domain of the function.
- We cancel the factor of x –2 and then put x = 2. So,

\(\Rightarrow\lim\limits_{x\rightarrow 2}\;\dfrac {x-2}{(x-2)(x-3)} \)

\(= \lim\limits_{x\rightarrow 2}\;\dfrac {1}{(x-3)}\)

\(=\dfrac {1}{2-3}=-1\)

- Remove \(\dfrac {0}{0}\) form and then use direct substitution rule.

The Greatest Integer Function does not have a Limit at all Integers

This happens because the values of one side limits are different at all integers.

e.g.

\(\lim\limits_{x\rightarrow 1^+}\;[[x]]=1\), whereas \(\lim\limits_{x\rightarrow 1^-}\;[[x]]=0\)

(1^{+} = 1.0001) (1^{–} = 0.9999)

A 5

B 7

C –6

D Limit doesn't exist

To factorize expression of the form \(ax^2 + bx + c\) ,we think of two numbers whose product is ac and sum is b. Then write b as sum of those two numbers.

e.g.

\(2x^2 + 9x -11\) ( sum = 9, product = – 22 \(\therefore\) numbers are 11,–2 )

\( = 2x^2 +11x - 2x - 11\)

\( = x (2x +11) - 1(2x +11)\)

\(=(x-1)(2x + 11) \)

A \((x-1)(2x+9)\)

B \((3x + 8)(x+1)\)

C \((3x + 1)(x+ 4)\)

D \((3x + 5)(2x-1)\)

Consider the limit,

\(\lim\limits_{x\rightarrow 0} \left ( \dfrac {\sqrt {1+x}-1}{x} \right)\)

If we put x = 0, we get \(\dfrac {0}{0}\) form, we rationalize the numerator.

\(\lim\limits_{x\rightarrow 0} \dfrac {\sqrt {1+x}-1}{x} \)\(× \dfrac {\sqrt {1+x}+1} {\sqrt {1+x}+1}\) (Divide & multiply by \({\sqrt {1+x}+1}\))

\(=\lim\limits_{x\rightarrow 0} \dfrac {{1+x}-1}{x(\sqrt {1+x}+1)} \)

\(=\lim\limits_{x\rightarrow 0} \dfrac {1}{(\sqrt {1+x}+1)}=\dfrac {1}{2} \)

- For rationalizing \(\sqrt a-\sqrt b\), we use the factor \(\sqrt a+\sqrt b\).

A \(\dfrac {1}{6}\)

B \(\dfrac {2}{3}\)

C \(\dfrac {7}{12}\)

D \(\dfrac {8}{13}\)

We know that,

\(\lim\limits_{x\rightarrow a}\;f(x)\) exists when,

\(\lim\limits_{x\rightarrow a^+}\;f(x) = \lim\limits_{x\rightarrow a^-}\;f(x)\)

\(\Rightarrow\) R.H.L. = L.H.L. = limit of the function.

- In some functions it is necessary to verify the above fact.

**\(f(x)=|x|\)**

** \(=x\) **if** \(x\geq0\) **

\(=-x\) if \(x<0\)

- If the quantity inside the modulus function is positive then we open the modulus.
- If the quantity inside the modulus function is negative then we open the modulus with negative sign.

A –4

B 10

C 16

D Limit does not exist

We define,

\(f(x)=[[x]]\) as greatest integer function, it is the value of greatest integer not greater than x.

For example

[[5.6]] = 5, [[8]] = 8, [[–3.2 ]] = –4, \([[\sqrt 3]]=1\)

The graph of this function is as shown below. The greatest integer function of any integer is the integer itself.

Due to the nature of the graph of this function, it is called step function. Observe that the graph is broken at all integral values of x.