Limits and Derivatives Class 11 Notes CBSE Maths Chapter 13 (Free PDF Download)

Section–A (1 Mark Questions)

1. If $\lim _{x \rightarrow 2} \frac{x^n-2^n}{x-2}=80$  and $n\epsilon N$ find n.

Ans. We have, $\lim _{x \rightarrow 2} \frac{x^n-2^n}{x-2}=80$

$$\begin{aligned}& \Rightarrow n \cdot 2^{n-1}=80 \\& \Rightarrow n \cdot 2^{n-1}=5 \cdot 2^{5-1} \\& \Rightarrow n=5 .\end{aligned}$$

2. $\lim _{x \rightarrow 1} \frac{x^2+1}{x+100}=\frac{1^2+1}{1+100} =?$

Ans. We have,

$$\lim _{x \rightarrow 1} \frac{x^2+1}{x+100}=\frac{1^2+1}{1+100}=\frac{2}{101}$$

3. If $f(x)=x^n$ and $if f{}'(1)=10$  find the value of n. 

Ans. We have, $f(x)=x^n$.

Differentiating both sides

w.r.t we get $f^{\prime}(x)=n x^{n-1}$.

Putting $x=1$ we get

$$f^{\prime}(1)=n \Rightarrow 10=n\left[\because f^{\prime}(1)=10\right]$$

4. Compute the derivative of $f(x)=6 x^{200}-x^{50}+x$. 

Ans. We have, $f(x)=6 x^{200}-x^{50}+x$

$$\begin{aligned}\Rightarrow f^{\prime}(x) & =6\left(200 x^{190}\right)-\left(50x^{69}\right)+1 \\& =1200 x^{199}-50 x^{49}+1\end{aligned}$$

5. Evaluate $\lim _{x \rightarrow 0} \frac{x}{\cos x}$ .

Ans. $\lim _{x \rightarrow 0} \frac{x}{\cos x}=\frac{0}{\cos 0}=\frac{0}{1}=0$

Section–B (2 Marks Questions)

6. Evaluate: $\lim _{x \rightarrow 2} \frac{x^3-4 x^2+4 x}{x^2-4}$ 

Ans. Evaluating the function at 2 , it is of the form $\frac{0}{0}$,

Hence,

$$\lim _{x \rightarrow 2} \frac{x^3-4 x^2+4 x}{x^2-4}=\lim _{x \rightarrow 2}\frac{x(x-2)^2}{(x+2)(x-2)}$$

$$\begin{gathered}=\lim _{x \rightarrow 2} \frac{x(x-2)}{(x+2)} \text { as } x \neq 2 \\=\frac{2(2-2)}{2+2}=\frac{0}{4}=0 .\end{gathered}$$

7. Evaluate: $\lim _{x \rightarrow 2} \frac{x^3-2 x^2}{x^2-5 x+6}$ 

Ans. Evaluating the function at 2, we get it of the form $\frac{0}{0}$,

Hence,

$$\begin{aligned}& \lim _{x \rightarrow 2} \frac{x^3-2 x^2}{x^2-5 x+6}=\lim _{x\rightarrow 2} \frac{x^2(x-2)}{(x-2)(x-3)} \\& =\lim _{x \rightarrow 2} \frac{x^2}{(x-3)}=\frac{(2)^2}{2-3}=\frac{4}{-1}=-4 .\end{aligned}$$

8. Find the derivative of $4 \sqrt{x}-2$ .

Ans. Let, $f(x)=4 \sqrt{x}-2$

$$\begin{aligned}& f^{\prime}(x)=\frac{d}{d x}(4 \sqrt{x}-2)=\frac{d}{d x}(4 \sqrt{x})-\frac{d}{d x}(2) \\& =4 \frac{d}{d x}\left(x^{\frac{1}{2}}\right)-0=4\left(\frac{1}{2} x^{\frac{1}{2}}\right) \\& =\left(2 x^{-\frac{1}{2}}\right)=\frac{2}{\sqrt{x}}\end{aligned}$$

9. Find the value of $lim_{x \rightarrow 2} \frac{x^2-4}{x^3-4 x^2+4 x}$ .

Ans. Evaluating the function at 2, we get it of the form $\frac{0}{0}$,

Hence,

$$\begin{aligned}& \lim _{x \rightarrow 2} \frac{x^2-4}{x^3-4 x^2+4 x}=\lim _{x\rightarrow 2} \frac{(x+2)(x-2)}{x(x-2)^2} \\& =\lim _{x \rightarrow 2} \frac{(x+2)}{x(x-2)}=\frac{2+2}{2(2-2)}=\frac{4}{0} \text { which is not }\end{aligned}$$

Defined.

10. Evaluate: $\lim_{x\rightarrow 0}\frac{\sqrt{1+x}-1}{x}$

Ans. Put $y=1+x$, so that $y \rightarrow 1$ as $x \rightarrow 0$.

Then, $\lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1}{x}=\lim _{y \rightarrow 1} \frac{\sqrt{y}-1}{y-1}$

$$\begin{aligned}& =\lim _{y \rightarrow 1} \frac{y^{\frac{1}{2}}-1^{\frac{1}{2}}}{y-1} \\& =\frac{1}{2}(1)^{\frac{1}{2}-1}=\frac{1}{2}\end{aligned}$$

11. Find $\lim_{x\rightarrow 1}f(x)$  where $f(x)= \begin{cases}x^2-1, & x \leq 1 \\-x^2-1, & x>1\end{cases}$ 

Ans. The given functions is

$$\begin{aligned}& f(x)= \begin{cases}x^2-1, & x \leq 1 \\-x^2-1, & x>1\end{cases} \\& \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}\left[x^2-1\right]=1^2-1=1-1=0 \\& \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}\left[-x^2-1\right]=-1^2-1=-1-1=-2\end{aligned}$$

It is observed that $\lim _{x \rightarrow \Gamma^{-}} f(x) \neq \lim _{x \rightarrow1^{-}} f(x)$.

Hence, $\lim _{x \rightarrow 1} f(x)$ does not exist.

12. Evaluate: $\lim _{x \rightarrow \dfrac{\pi}{6}} \dfrac{2 \sin ^2 x+\sin x-1}{2 \sin ^2 x-3 \sin x+1}$ 

Ans. $$ \begin{aligned} & \text { } \lim _{x \rightarrow \frac{\pi}{6}} \frac{2 \sin ^2 x+\sin x-1}{2 \sin ^2 x-3 \sin x+1} \\ & =\lim _{x \rightarrow \frac{\pi}{6}} \frac{(2 \sin x-1)(\sin x+1)}{(2 \sin x-1)(\sin x-1)}=\lim _{x \rightarrow \frac{\pi}{6}} \frac{\sin x+1}{\sin x-1} \\ & =\frac{\frac{1}{2}+1}{\frac{1}{2}-1}=-3 \end{aligned} $$

PDF Summary - Class 11 Maths Limits and Derivatives (Chapter 13)

Limits: 

• Consider a function $f(x)={{x}^{2}}$. Plotting it gives

Here, the value of x approaches 0 as the value of function f(x) moves to 0.

• In general as \[x \to a,\text{ }f\left( x \right)\to l\] , then \[l\] is called limit of the function \[f\left( x \right)\] . This is written symbolically as \[\displaystyle \lim_{x \to a}f(x)=l\].

• Irrespective of the limits, the function should assume at a given point \[x=a\].

• There are two ways in which $x$ can approach a number. It can either be from left or from right. This means that all the values of $x$ near $a$ could be less than $a$ or could be greater than $a$.

• Right hand limit - Value of \[f\left( x \right)\] which is dictated by values of \[f\left( x \right)\] when $x$ tends to from the right. It is written as $\displaystyle \lim_{x \to {{a}^{+}}}f(x)$.

• Left hand limit - Value of \[f\left( x \right)\] which is dictated by values of \[f\left( x \right)\] when $x$ tends to from the left. It is written as $\displaystyle \lim_{x \to {{a}^{-}}}f(x)$.

• Here, the right and left hand limits are different. So, the limit of \[f\left( x \right)\] as \[x\] tends to zero does not exist (even though the function is defined at \[0\]).

• If the right and left hand limits coincide then the common value is the limit and denoted by $\displaystyle \lim_{x \to a}f(x)$.

Algebra of limits:

Theorem 1:

Let $f$ and $g$ be two functions such that both \[\displaystyle \lim_{x \to a}f(x)\text{ and }\displaystyle \lim_{x \to a}g(x)\] exist, then 

• Limit of sum of two functions is sum of the limits of the functions, i.e. 

\[\displaystyle \lim_{x \to a}\left[ f(x)+g(x) \right]=\displaystyle \lim_{x \to a}f(x)+\displaystyle \lim_{x \to a}g(x)\] .

• Limit of difference of two functions is difference of the limits of the functions, i.e. 

\[\displaystyle \lim_{x \to a}\left[ f(x)-g(x) \right]=\displaystyle \lim_{x \to a}f(x)-\displaystyle \lim_{x \to a}g(x)\]

• Limit of product of two functions is product of the limits of the functions, i.e., 

\[\displaystyle \lim_{x \to a}\left[ f(x).g(x) \right]=\displaystyle \lim_{x \to a}f(x).\displaystyle \lim_{x \to a}g(x)\]

• Limit of quotient of two functions is quotient of the limits of the functions (whenever the denominator is non zero), i.e., 

\[\displaystyle \lim_{x \to a}\dfrac{f(x)}{g(x)}=\dfrac{\displaystyle \lim_{x \to a}f(x)}{\displaystyle \lim_{x \to a}g(x)}\]

• In particular as a special case of (iii), when g is the constant function such that \[g\left( x \right)=\lambda \] , for some real number \[\lambda \] , we have 

\[\displaystyle \lim_{x \to a}\left[ \left( \lambda .f \right)(x) \right]=\lambda .\displaystyle \lim_{x \to a}f(x)\]

Limits of polynomials and rational functions:

• A function \[f\] is said to be a polynomial function if \[f\left( x \right)\] is zero function or if \[f\left(x\right)={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+...+{{a}_{n}}{{x}^{n}}\] where ${{a}_{i}}S$ is are real numbers such that ${{a}_{n}}\ne 0$ for some natural number $n$.

• We know that \[\displaystyle \lim_{x \to a}x=a\]

\[\displaystyle \lim_{x \to a}{{x}^{2}}=\displaystyle \lim_{x \to a}\left( x.x \right)=\displaystyle \lim_{x \to a}x.\displaystyle \lim_{x \to a}x=a.a={{a}^{2}}\] 

Hence, 

\[\displaystyle \lim_{x \to a}{{x}^{n}}={{a}^{n}}\] 

• Let \[f\left( x \right)={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+...+{{a}_{n}}{{x}^{n}}\] be a polynomial function 

\[\displaystyle \lim_{x \to a}f\left( x \right)=\displaystyle \lim_{x \to a}\left[ {{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+...+{{a}_{n}}{{x}^{n}} \right]\]

$=\displaystyle \lim_{x \to a}{{a}_{0}}+\displaystyle \lim_{x \to a}{{a}_{1}}x+\displaystyle \lim_{x \to a}{{a}_{2}}{{x}^{2}}+...+\displaystyle \lim_{x \to a}{{a}_{n}}{{x}^{n}}$ 

$={{a}_{0}}+{{a}_{1}}\displaystyle \lim_{x \to a}x+{{a}_{2}}\displaystyle \lim_{x \to a}{{x}^{2}}+...+{{a}_{n}}\displaystyle \lim_{x \to a}{{x}^{n}}$ 

\[={{a}_{0}}+{{a}_{1}}a+{{a}_{2}}{{a}^{2}}+...+{{a}_{n}}{{a}^{n}}\] 

$=f(a)$ 

• A function $f$ is said to be a rational function, if $f(x)=\dfrac{g(x)}{h(x)}$ where \[g\left( x \right)\text{ and }h\left( x \right)\] are polynomials such that \[h\left( x \right)\ne 0\] . 

Then

$\displaystyle \lim_{x \to a}f(x)=\displaystyle \lim_{x \to a}\dfrac{g(x)}{h(x)}=\dfrac{\displaystyle \lim_{x \to a}g(x)}{\displaystyle \lim_{x \to a}h(x)}=\dfrac{g(a)}{h(a)}$  

• However, if $h(a)=0$ , there are two scenarios – 

• when \[g\left( a \right)\ne 0\] 

• limit does not exist 

• When $g(a)=0$ . 

• \[g\left( x \right)={{\left( xa \right)}^{k}}{{g}_{1}}\left( x \right)\] , where $k$ is the maximum of powers of $\left( x-a \right)$ in $g(x)$ . 

• Similarly, \[h\left( x \right)={{\left( xa \right)}^{l}}{{h}_{1}}\left( x \right)\] as $h(a)=0$ . Now, if \[k\ge l\] , we have

$\displaystyle \lim_{x \to a}f(x)=\dfrac{\displaystyle \lim_{x \to a}g(x)}{\displaystyle \lim_{x \to a}h(x)}=\dfrac{\displaystyle \lim_{x \to a}{{\left( xa \right)}^{k}}{{g}_{1}}\left( x \right)}{\displaystyle \lim_{x \to a}{{\left( xa \right)}^{l}}{{h}_{1}}\left( x \right)}$

\[=\dfrac{\displaystyle \lim_{x \to a}{{\left( xa \right)}^{\left( k-l \right)}}{{g}_{1}}\left( x \right)}{\displaystyle \lim_{x \to a}{{h}_{1}}\left( x \right)}=\dfrac{0.{{g}_{1}}(a)}{{{h}_{1}}(a)}=0\] 

 If \[k < l\] , the limit is not defined. 

Theorem 2:

For any positive integer $n$, $\displaystyle \lim_{x \to a}\dfrac{{{x}^{n}}-{{a}^{n}}}{x-a}=n{{a}^{n-1}}$.

The proof is shown below.

Dividing $\left( {{x}^{n}}-{{a}^{n}} \right)$ by $\left( x-a \right)$,

$\displaystyle \lim_{x \to a}\dfrac{{{x}^{n}}-{{a}^{n}}}{x-a}=\displaystyle \lim_{x \to a}\left( {{x}^{n-1}}+{{x}^{n-2}}a+{{x}^{n-3}}{{a}^{2}}+...+x{{a}^{n-2}}+{{a}^{n-1}} \right)$ 

$={{a}^{n-1}}+a\,{{a}^{n-2}}+...+{{a}^{n-2}}(a)+{{a}^{n-1}}$ 

$={{a}^{n-1}}+{{a}^{n-1}}+...+{{a}^{n-1}}+{{a}^{n-1}}\,\left( n\text{ terms} \right)$ 

$=n{{a}^{n-1}}$ 

The expression in the above theorem for the limit is true even if $n$ is any rational number and $a$ is positive. 

Limits of Trigonometric Functions:

Theorem 3:

Let $f$ and $g$ be two real valued functions with the same domain such that \[f\left( x \right)\le g\left( x \right)\] for all $x$ in the domain of definition, 

For some $a$ , if both $\displaystyle \lim_{x \to a}f(x)\text{ and }\displaystyle \lim_{x \to a}g(x)$ exist, then $\displaystyle \lim_{x \to a}f(x)\le \displaystyle \lim_{x \to a}g(x)$

         

Theorem 4 (Sandwich Theorem):

Let $f,g$ and $h$ be real functions such that \[f\left( x \right)\le g\left( x \right)\le h\left( x \right)\] for all $x$ in the common domain of definition. 

For some real number $a,\text{ if }\displaystyle \lim_{x \to a}f(x)=l=\displaystyle \lim_{x \to a}g(x)$ , then $\displaystyle \lim_{x \to a}g(x)=l$ .

To prove:

$\cos x < \dfrac{\sin x}{x} < 1$  for $0 < \left| x \right| < \dfrac{\pi }{2}$ 

Proof: Use known facts that \[\sin \left( \text{ }x \right)=\sin x\] and \[\cos \left( \text{ }x \right)=\cos x\]. Hence, it is sufficient to prove the inequality for $0 < x < \dfrac{\pi }{2}$

• From the figure, it is noted that O is the centre of the unit circle such that the angle \[AOC\] is $x$ radians and $0 < x < \dfrac{\pi }{2}$ . 

• Two perpendiculars to \[OA\] are the line segments \[BA\] and \[CD\]. 

• Now join \[AC\] and then,

\[\text{Area of }\Delta OAC < \text{Area of sector }OAC < \text{Area of }\Delta OAB\] 

i.e., $\dfrac{1}{2}OA.CD < \dfrac{x}{2\pi }.\pi .{{\left( OA \right)}^{2}} < \dfrac{1}{2}OA.AB$ 

i.e., $CD < x.OA < AB$ 

From $\Delta OCD$,

$\sin x=\dfrac{CD}{OA}$ (since $OC=OA$) and hence $CD=OA\sin x$. 

Also $\tan x=\dfrac{AB}{OA}$ and hence $AB=OA\tan x$. Thus

$OA\sin x < OA.x < OA\tan x$ 

Since $OA$ is positive, 

$\sin x < x < \tan x$ 

Since $0 < x < \dfrac{\pi }{2}$, $\sin x$ is positive and thus by dividing throughout by $\sin x$,

$1 < \dfrac{x}{\sin x} < \dfrac{1}{\cos x}$. 

Taking reciprocals throughout,

$\sin x < x < \tan x$ .

Since $0 < x < \dfrac{\pi }{2}$, $\sin x$ is positive and thus by dividing throughout by $\sin x$,

$1 < \dfrac{x}{\sin x} < \dfrac{1}{\cos x}$. 

Taking reciprocals throughout, 

$\cos x < \dfrac{\sin x}{x} < 1$ 

Hence, Proved.

The following are two important limits 

(i). $\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=1$  

(ii). $\displaystyle \lim_{x \to 0}\dfrac{1-\cos x}{x}=0$ 

The proof is given as below,

The function $\dfrac{\sin x}{x}$ is sandwiched between the function $\cos x$ and the constant function which takes value $1$.

Since $\displaystyle \lim_{x \to 0}\cos x=1$ and $1-\cos x=2{{\sin }^{2}}\left( \dfrac{x}{2} \right)$,

$\displaystyle \lim_{x \to 0}\dfrac{1-\cos x}{x}=\displaystyle \lim_{x \to 0}\dfrac{2{{\sin }^{2}}\left( \dfrac{x}{2} \right)}{x}=\displaystyle \lim_{x \to 0}\dfrac{\sin \left( \dfrac{x}{2} \right)}{\dfrac{x}{2}}.\sin \left( \dfrac{x}{2} \right)$

$=\displaystyle \lim_{x \to 0}\dfrac{{{\sin }^{2}}\left( \dfrac{x}{2} \right)}{\dfrac{x}{2}}.\displaystyle \lim_{x \to 0}\sin \left( \dfrac{x}{2} \right)=1.0=0$

Using the fact that \[x \to 0\] is equivalent to $\dfrac{x}{2}\to 0$. This may be justified by putting $y=\dfrac{x}{2}$.

Derivatives:

Derivative of a function at a given point in its domain of definition. 

• Definition 1 - Suppose $f$ is a real valued function and $a$ is a point in its domain of definition. The derivative of $f$ at a is defined by $\displaystyle \lim_{h\to 0}\dfrac{f\left( a+h \right)-f\left( a \right)}{h}$, provided this limit exists. ${f}'\left( a \right)$ is used to denoted the derivative of $f\left( x \right)$ at a.

• Definition 2 - Suppose $f$ is a real valued function, the function defined by $\displaystyle \lim_{h\to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$ wherever limit exists is defined to be derivative of $f$ at $x$ denoted by ${f}'\left( x \right)$. This definition of derivative is also called the first principle of derivative. 

Thus ${f}'\left( x \right)=\displaystyle \lim_{h\to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$.

The derivative of function f(x) with respect to x can be denoted in two ways:${f}'\left( x \right)$ is denoted by $\dfrac{d}{dx}\left( f\left( x \right) \right)$ or if $y=f\left( x \right)$, it is denoted by $\dfrac{dy}{dx}$. 

Another notation is $D\left( f\left( x \right) \right)$.

Further, derivative of f at $x=a$ is also denoted by \[{{\left. \dfrac{d}{dx}f\left( x \right) \right|}_{a}}\text{ or }{{\left. \dfrac{df}{dx} \right|}_{a}}\text{ or even }{{\left( \dfrac{df}{dx} \right)}_{x=a}}\].

Theorem 5:

• Let $f$ and $g$ be two functions such that their derivatives are defined in a common domain. Then 

• Derivative of sum of two functions is sum of the derivatives of the functions. 

\[\dfrac{d}{dx}\left[ f\left( x \right)+g\left( x \right) \right]=\dfrac{d}{dx}f\left( x \right)+\dfrac{d}{dx}g\left( x \right)\]

• Derivative of difference of two functions is difference of the derivatives of the functions.

\[\dfrac{d}{dx}\left[ f\left( x \right)-g\left( x \right) \right]=\dfrac{d}{dx}f\left( x \right)-\dfrac{d}{dx}g\left( x \right)\]

• Derivative of product of two functions is given by following product rule. 

\[\dfrac{d}{dx}\left[ f\left( x \right).g\left( x \right) \right]=\dfrac{d}{dx}f\left( x \right).g(x)+f(x).\dfrac{d}{dx}g\left( x \right)\]

• Derivative of quotient of two functions is given by the following quotient rule (whenever the denominator is non–zero).

\[\dfrac{d}{dx}\left( \dfrac{f(x)}{g(x)} \right)=\dfrac{\dfrac{d}{dx}f(x).g(x)-f(x)\dfrac{d}{dx}g(x)}{{{\left( g(x) \right)}^{2}}}\] 

• Let $u=f(x)$ and $v=g(x)$ . 

• Product Rule:

• ${{\left( uv \right)}^{\prime }}={u}'v+u{v}'$ . 

• Also referred as Leibnitz rule for differentiating product of functions 

• Quotient Rule:

• ${{\left( \dfrac{u}{v} \right)}^{\prime }}=\dfrac{{u}'v-u{v}'}{{{v}^{2}}}$

• Derivative of the function $f(x)=x$ is the constant.

Theorem 6:

Derivative of \[f(x)={{x}^{n}}\] is \[n{{x}^{n-1}}\] for any positive integer \[n\]. 

Proof 

• By definition of the derivative function, we have 

${f}'(x)=\displaystyle \lim_{h\to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}=\displaystyle \lim_{h\to 0}\dfrac{{{\left( x+h \right)}^{n}}-{{x}^{n}}}{h}$ .

Binomial theorem tells that ${{\left( x+h \right)}^{n}}=\left( {}^{n}{{C}_{0}} \right){{x}^{n}}+\left( {}^{n}{{C}_{1}} \right){{x}^{n-1}}h+...+\left( {}^{n}{{C}_{n}} \right){{h}^{n}}$ and ${{\left( x+h \right)}^{n}}-{{x}^{n}}=h\left( n{{x}^{n-1}}+...+{{h}^{n-1}} \right)$. Thus

$\dfrac{df(x)}{dx}=\displaystyle \lim_{h\to 0}\dfrac{{{\left( x+h \right)}^{n}}-{{x}^{n}}}{h}$ 

$=\displaystyle \lim_{h\to 0}\dfrac{h\left( n{{x}^{n-1}}+...+{{h}^{n-1}} \right)}{h}$ 

$=\displaystyle \lim_{h\to 0}\left( n{{x}^{n-1}}+...+{{h}^{n-1}} \right)=n{{x}^{n-1}}$ 

• This can be proved as below alternatively

\[\dfrac{d}{dx}\left( {{x}^{n}} \right)=\dfrac{d}{dx}\left( x.{{x}^{n-1}} \right)\] 

\[=\dfrac{d}{dx}(x).\left( {{x}^{n-1}} \right)+x.\dfrac{d}{dx}\left( {{x}^{n-1}} \right)\]  (By product rule)

$=1.{{x}^{n-1}}+x.\left( (n-1){{x}^{n-2}} \right)$ (By induction hypothesis)

$={{x}^{n-1}}+(n-1){{x}^{n-1}}=n{{x}^{n-1}}$ 

Theorem 7:

• Let $f(x)={{a}_{n}}{{x}^{n}}+{{a}_{n-1}}{{x}^{n-1}}+....+{{a}_{1}}x+{{a}_{0}}$ be a polynomial function, where ${{a}_{i}}$ s are all real numbers and ${{a}_{n}}\ne 0$. Then, the derivative function is given by 

$\dfrac{df(x)}{dx}=n{{a}_{n}}{{x}^{n-1}}+(n-1){{a}_{n-1}}{{x}^{n-2}}+....+2{{a}_{2}}x+{{a}_{1}}$ 

Quick Reference:

• For functions $f$ and $g$ the following holds: 

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

\[\displaystyle \lim_{x \to a}\left[ f(x).g(x) \right]=\displaystyle \lim_{x \to a}f(x).\displaystyle \lim_{x \to a}g(x)\]

\[\displaystyle \lim_{x \to a}\dfrac{f(x)}{g(x)}=\dfrac{\displaystyle \lim_{x \to a}f(x)}{\displaystyle \lim_{x \to a}g(x)}\]

• Following are some of the standard limits 

$\displaystyle \lim_{x \to a}\dfrac{{{x}^{n}}-{{a}^{n}}}{x-a}=n{{a}^{n-1}}$

$\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=1,\displaystyle \lim_{x \to a}\dfrac{\sin \left( x-a \right)}{x-a}=1$

$\displaystyle \lim_{x \to 0}\dfrac{1-\cos x}{x}=0$

$\displaystyle \lim_{x \to 0}\dfrac{\tan x}{x}=1,\displaystyle \lim_{x \to a}\dfrac{\tan \left( x-a \right)}{x-a}=1$

$\displaystyle \lim_{x \to 0}\dfrac{{{\sin }^{-1}}x}{x}=1,\displaystyle \lim_{x \to 0}\dfrac{{{\tan }^{-1}}x}{x}=1$

$\displaystyle \lim_{x \to 0}\dfrac{{{a}^{x}}-1}{x}={{\log }_{e}}a,a > 0,a\ne 1$

• Derivatives 

• The derivative of a function $f$ at $a$ is defined by

${f}'\left( a \right)=\displaystyle \lim_{h\to 0}\dfrac{f\left( a+h \right)-f\left( a \right)}{h}$

• Derivative of a function $f$ at any point $x$ is defined by

${f}'\left( x \right)=\dfrac{df(x)}{dx}=\displaystyle \lim_{h\to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$ 

• For functions $u$ and $v$ the following holds: 

${{\left( u\pm v \right)}^{\prime }}={u}'\pm {v}'$

${{\left( uv \right)}^{\prime }}={u}'v+u{v}'\Rightarrow \dfrac{d}{dx}\left( uv \right)=u.\dfrac{dv}{dx}+v.\dfrac{du}{dx}$

${{\left( \dfrac{u}{v} \right)}^{\prime }}=\dfrac{{u}'v-u{v}'}{{{v}^{2}}}\Rightarrow \dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v.\dfrac{du}{dx}-u.\dfrac{dv}{dx}}{{{v}^{2}}}$ 

• Following are some of the standard derivatives

\[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\]

\[\dfrac{d}{dx}\left( \sin x \right)=\cos x\]

\[\dfrac{d}{dx}\left( \cos x \right)=-\sin x\]

\[\dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x\]

\[\dfrac{d}{dx}\left( \cot x \right)=-\text{cose}{{\text{c}}^{2}}x\]

\[\dfrac{d}{dx}\left( \sec x \right)=\sec x.\tan x\]

\[\dfrac{d}{dx}\left( \cos \text{ec }x \right)=-\cos \text{ec }x.\cot x\]

Limits and Derivatives Class 11 Notes – Chapter Overview

Class 11 Revision Notes Limits and Derivatives PDF 

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Class 11 Maths Revision Notes Chapter 13 

Limits - In Calculus, we study how the value of a function changes with a change in the points in the domain. The limit of a function is stated as x -> p, f(x) -> 1, where 1 is the limit of the function f(x). Its notation is: x pf(x) = l. Some key points about the limit of a function are:

• X could approach a number from either the left or right, i.e., all the values that x could assume near p could be either less than or greater than p.

• Limits are of two types:

a. Right-hand limit - When x tends to p from the right, then that f(x) value is the right-hand limit.

b. Left-hand limit - When x tends to p from the left, then that value of f(x) is the left-hand limit.

Standard Limits - Listed Below Are Some Standard Limits

1.  \[\lim_{x \to p}\] (xn - pn)/x - p = npn-1

2.  \[\lim_{x \to 0}\] (sin x)/x = 1

3. \[\lim_{x \to p}\] (sin (x - p))/(x - p) = 1

4. \[\lim_{x \to 0}\] (tan x)/x = 1

5. \[\lim_{x \to p}\] (tan (x - p))/(x - p) = 1

6. \[\lim_{x \to 0}\] (sin-1 x)/x = 1

7. \[\lim_{x \to 0}\] (tan-1 x)/x = 1

8. \[\lim_{x \to 0}\] (px - 1)/x = logep, p > 0 and p 1

Algebra of Limits - If f and z are two functions and limits for both exists  (\[\lim_{x \to p}\]f(x), \[\lim_{x \to p}\]z(x)), then:

• The limit of the sum of two functions is equal to the individual sum of limits of the two functions.

\[\lim_{x \to p}\] [f(x) + z(x)] = \[\lim_{x \to p}\]f(x) + \[\lim_{x \to p}\]z(x)

• The limit of the difference of two functions is equal to the difference in limits of the two functions.

\[\lim_{x \to p}\] [f(x) - z(x)] = \[\lim_{x \to p}\]f(x) - \[\lim_{x \to p}\]z(x)

• The limit of the multiplication of two functions is equal to the product of the limits of the two functions.

\[\lim_{x \to p}\] [f(x) * z(x)] = \[\lim_{x \to p}\]f(x) * \[\lim_{x \to p}\]z(x)

• If z is c constant function and z(x) = λ then:

\[\lim_{x \to p}\] [(λ * f)(x)] =  λ * \[\lim_{x \to p}\]f(x)

• The limit of the division of two functions is equal to the division of limits of the two functions, provided the denominator is non-zero.

\[\lim_{x \to p}\][f(x)/z(x)] = \[\lim_{x \to p}\]f(x)/\[\lim_{x \to p}\]z(x)

Derivatives - Derivative of a function y = f(x) is found by using the formulas change in y/change in x. Lets us say x changes from x to dx, then y changes from y to f(x) to f(x + dx) so:

• Derivative = change in y/change in x = dy/dx = f(x + dx) - f(x)/dx

• Derivative of a function at a point p is given by:

f’(p) = \[\lim_{h \to 0}\]f(p + h)/h

Some Standard Derivatives - Mentioned Below Are Some Standard Derivatives

1. d(xi)/dx = ixi-1

2. d(sin x)/dx = cos x

3. d(cos x)/dx = -sin x

4. d(tan x)/dx = sec2x

5. d(cot x)/dx = -coses2x

6. d(sec x)/dx = secx . tanx

7. d(cosex x)/dx = -cosec x. cot x