Limits and Derivatives

Class 11 NCERT

NCERT

1   Evaluate the Given limit:$\lim \limits_{x \to 3}x+3$

Solution :

$\lim \limits_{x \to 3}x+3=3+3=6$

2   Evaluate the Given limit:$\lim \limits_{x \to \pi} (x-\dfrac{22}{7})$

Solution :

$\lim \limits_{x \to \pi}(x-\dfrac{22}{7})=(\pi-\dfrac{22}{7})$

3   Evaluate the Given limit:$\lim \limits_{r \to 1}\pi r^2$

Solution :

$\lim \limits_{r \to 1} \pi r^2=\pi(1^2)=\pi$

4   Evaluate the Given limit: $ \lim \limits_{x \to 1}\dfrac{4x+3}{x-2}$

Solution :

$\lim \limits_{x \to 1} \dfrac{4x+3}{x-2}=\dfrac{4(1)+3}{1-2}=\dfrac{4+3}{-1}=-7$

5   Evaluate the Given limit: $ \lim \limits_{x \to -1}\dfrac{x^{10}+x^5+1}{x-1}$

Solution :

$\lim \limits_{x \to -1}\dfrac{x^{10}+x^5+1}{x-1}\\ =\dfrac{(-1)^{10}+(-1)^5+1}{-1-1}\\ =\dfrac{1-1+1}{-2}=-\dfrac{1}{2}$

6   Evaluate the Given limit: $ \lim\limits_{x \to 0}\dfrac{(x+1)^5-1}{x}$

Solution :

$\lim\limits_{x \to 0}\dfrac{(x+1)^5-1}{x}$Put $x+1=y$ so that $ y \to 1 $ as $ x\to 0$$\\$ Accordingly,$ \lim\limits_{x \to 0}\dfrac{(x+1)^5-1}{x}= \lim\limits_{x \to 1}\dfrac{(y)^5-1}{y-1}$$\\$ $=5.1^{5-1}[\lim\limits_{x \to a}\dfrac{x^n- a^n}{x-a}=na^{n-1}]\\ =5\\ \therefore \lim\limits_{x \to 0}\dfrac{(x+1)^5-1}{x}=5$

7   Evaluate the Given limit: $ \lim\limits_{x \to 2} \dfrac{3x^2-x-10}{x^2-4}$

Solution :

At $x = 2$, the value of the given rational function takes the form $\dfrac{0}{0}$ $\therefore \lim\limits_{x \to 2} \dfrac{3x^2-x-10}{x^2-4}\\ =\lim\limits_{x \to 2}\dfrac{(x-2)(3x+5)}{(x-2)(x+2)}\\ =\lim\limits_{x \to 2}\dfrac{3x+5}{x+2}\\ =\dfrac{3(2)+5}{2+2}\\ =\dfrac{11}{4}$

8   Evaluate the Given limit: $\lim\limits_{x \to 3}\dfrac{x^4-81}{2x^2-5x-3}$

Solution :

At $x = 2$, the value of the given rational function takes the form $\dfrac{0}{0}$$\\$ $\therefore \lim\limits_{x \to 3 }\dfrac{x64-81}{2x62-5x-3}\\ =\lim\limits_{x \to 3}\dfrac{(x-3)(x+3)(x^2+9)}{(x-3)(2x+1)}\\ =\lim\limits_{x \to 3}\dfrac{(x+3)(x^2+9)}{(2x+1)}\\ =\dfrac{(3+3)(3^2+9)}{2(3)+1}\\ =\dfrac{6*18}{7}\\ \dfrac{108}{7}$$\\$

9   Evaluate the Given limit:$\lim\limits_{ x\to 0}\dfrac{ax+b}{cx+1}$

Solution :

$\lim\limits_{ x\to 0}\dfrac{ax+b}{cx+1}$=$\dfrac{a(0)+b}{c(0)+1}=b$

10   Evaluate the Given limit: $ \lim\limits_{x \to 1}\dfrac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}$

Solution :

$ \lim\limits_{x \to 1}\dfrac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}$ At $z = 1$, the value of the given function takes the form $\dfrac{0}{0}$$\\$ Put $ z^{\frac{1}{6}} =x $ so that $ z\to 1$ as $ x \to 1.$$\\$ accordingly,$ \lim\limits_{z \to 1}\dfrac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}=\lim\limits_{x \to 1}\dfrac{x^2-1}{x-1}$$\\$ $=\lim\limits_{x \to 1}\dfrac{x^2-1}{x-1}\\ =2.1^{2-1} \quad [\lim\limits_{x \to a}\dfrac{x^n-a^n}{x-a}=na^{n-1}]\\ =2\\ \lim\limits_{x \to 1}\dfrac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}=2$