mathematics-class-12-unit-05-chapter-01-continuity-and-differentiability-l-1-xuha2cvlanu

### <Success style="font-size:25px"> Continuity And Differentiability L-1 </Success>
### <primary style="font-size:24px"> Continuity of a function at a point </primary>
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- Let $f$ be a real valued function defined on a domain $D \subseteq \mathbb{R}$

- Suppose $a \in D$ (ie. $f(a)$ is defined)

- We say that $f$ is continuous at a

- if $\quad f(a)=\lim _{x \rightarrow a} f(x)$
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<footer style = "font-size: 18px"> $\rightarrow$ Continuity & Differentiability Lecture -1 $\rightarrow$ <span style = "color:blue"> Continuity of a function at a point </span> $\rightarrow$ Continuous functions $\rightarrow$ Example 1 </footer>

### <Success style="font-size:25px"> Continuity And Differentiability L-1 </Success>
### <primary style="font-size:24px"> Continuous functions </primary>
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- Recall that for $\lim _{x \rightarrow a} f(x)$ to exist the function $f(x)$ must be defined in some open interval containing $a$.

- (but not necessarily at x=a)

- So, $f(x)$ in continuous at $x=a$ 
 
- (i) if $\lim _{x \rightarrow a} f(x)$ exists
- and (ii) $f(a)=\lim _{x \rightarrow a} f(x)$.

- So, for $f(x)$ to be continuous at $x=a$

- the  function must be defined at $x=a$

- and $f(a)$ must equal $\lim _{x \rightarrow a} f(x)$.  
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<footer style = "font-size: 18px">Continuity & Differentiability Lecture -1 $\rightarrow$ Continuity of a function at a point $\rightarrow$ <span style = "color:blue"> Continuous functions </span> $\rightarrow$ Example 1 $\rightarrow$ Example 1 </footer>

### <Success style="font-size:25px"> Continuity And Differentiability L-1 </Success>
### <primary style="font-size:24px"> Example 1 </primary>
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- f(x)= $\begin{cases}0, & x<0 \\\ 1, & x \geqslant 0\end{cases}$

- Here, $\lim _{x \rightarrow 0} f(x)$ does not
- exist (because $\lim _{x \rightarrow 0^{-}} f(x)=0 \neq 1=\lim _{x \rightarrow 0^{+}} f(x)$ )

- So, $f(x)$ is not continuous at $x=0$ (even though $f(x)$ is defined at $x=0 ; f(0)=1$ )
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<footer style = "font-size: 18px">Continuity of a function at a point $\rightarrow$ Continuous functions $\rightarrow$ <span style = "color:blue"> Example 1 </span> $\rightarrow$ Example 1 $\rightarrow$ Intuitive meaning of continuity </footer>

### <Success style="font-size:25px"> Continuity And Differentiability L-1 </Success>
### <primary style="font-size:24px"> Example 1 </primary>
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- But if $a \neq 0$, then $\lim _{x \rightarrow a} f(x)= \begin{cases}1, & \text { if } a>0 \\\ 0, & \text { if } a<0\end{cases}$

- Also, $f(a)= \begin{cases}1 & \text { if } a>0 \\\  0 & \text { if } a<0\end{cases}$

- $\therefore f(x)$ is continuous at all points except $x=0$.

- If $f(x)$ is not continuous at $x=a$, we will say that $f(x)$ is discontinuous at $x=a$.
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<footer style = "font-size: 18px">Continuous functions $\rightarrow$ Example 1 $\rightarrow$ <span style = "color:blue"> Example 1 </span> $\rightarrow$ Intuitive meaning of continuity $\rightarrow$ Piecewise function </footer>

### <Success style="font-size:25px"> Continuity And Differentiability L-1 </Success>
### <primary style="font-size:24px"> Intuitive meaning of continuity </primary>
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- If we can draw the graph of the function $y=f(x)$, then the continuity of $f(x)$ at $x=a$ means that the graph is not "broken" near the point $(a, f(a))$ in the graph.

- For $f(x)= \begin{cases}1, & x \geqslant 0 \\\ 0, & x<0\end{cases}$

- $f(x)$ has "broken" graph near $(0,1)$.
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<footer style = "font-size: 18px">Example 1 $\rightarrow$ Example 1 $\rightarrow$ <span style = "color:blue"> Intuitive meaning of continuity </span> $\rightarrow$ Piecewise function $\rightarrow$ Example 2 </footer>

### <Success style="font-size:25px"> Continuity And Differentiability L-1 </Success>
### <primary style="font-size:24px"> Piecewise function </primary>
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- (2) $\quad f(x)= \begin{cases}1, & x \neq 0 \\\ 0, & x=0\end{cases}$

- Here $\lim _{x \rightarrow 0} f(x)$ exists & is equal to 1 .

- But, $f(0)=0 \neq \lim _{x \rightarrow 0} f(x)$

- Hence, $f(x)$ is discontinuous  at $x=0$
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<footer style = "font-size: 18px">Example 1 $\rightarrow$ Intuitive meaning of continuity $\rightarrow$ <span style = "color:blue"> Piecewise function </span> $\rightarrow$ Example 2 $\rightarrow$ Example 2 </footer>

### <Success style="font-size:25px"> Continuity And Differentiability L-1 </Success>
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- Let's consider

- $\quad f(x)= \begin{cases}x \sin \left(\frac{1}{x}\right), & x \neq 0 \\\ 0, & x=0\end{cases}$

- Is $f$ continuous at $x=0$ ? It is difficult to draw the graph.

- But the $\lim _{x \rightarrow 0} f(x)$ can be calculated as follows.

- $|f(x)|=\left|x \sin \left(\frac{1}{x}\right)\right| \leqslant|x|\left|\sin \left(\frac{1}{x}\right)\right|$

- Since, $-1 \leqslant \sin \left(\frac{1}{x}\right) \leqslant 1$ for all $x \neq 0$.
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<footer style = "font-size: 18px">Intuitive meaning of continuity $\rightarrow$ Piecewise function $\rightarrow$ <span style = "color:blue"> Example 2 </span> $\rightarrow$ Example 2 $\rightarrow$ Differentiability of a function at a point </footer>

### <Success style="font-size:25px"> Continuity And Differentiability L-1 </Success>
### <primary style="font-size:24px"> Example 2 </primary>
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- $\therefore \quad\left|\sin \left(\frac{1}{x}\right)\right| \leqslant 1$

- $\therefore 0 \leqslant\left|x \sin \left(\frac{1}{x}\right)\right| \leqslant|x|$

- Since $\lim _{x \rightarrow 0} x=0=\lim _{x \rightarrow 0} 0$, by the Sandwich theorem, $\quad \lim _{x \rightarrow 0}\left|x \sin \left(\frac{1}{x}\right)\right|=0$

- $\Rightarrow \quad \lim _{x \rightarrow 0} | x \sin \left(\frac{1}{x}\right)|=0 $

- $\therefore \quad  f(0)=0=\lim _{x \rightarrow 0} f(x)$

- So, $f(x)$ is continuous at $x=0$.
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<footer style = "font-size: 18px">Piecewise function $\rightarrow$ Example 2 $\rightarrow$ <span style = "color:blue"> Example 2 </span> $\rightarrow$ Differentiability of a function at a point $\rightarrow$ Geometric interpretation </footer>

### <Success style="font-size:25px"> Continuity And Differentiability L-1 </Success>
### <primary style="font-size:24px"> Differentiability of a function at a point </primary>
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- Let $f(x)$ be a real-valued function.

- we say that $f(x)$ is differentiable at $x=a$ if the limit $\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$ exists.

- Note that for $f(x)$ to be differentiable at $x=a$ the function $f(x)$ must be defined in an open interval containing $x=a$.
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<footer style = "font-size: 18px">Example 2 $\rightarrow$ Example 2 $\rightarrow$ <span style = "color:blue"> Differentiability of a function at a point </span> $\rightarrow$ Geometric interpretation $\rightarrow$ Geometric interpretation </footer>

### <Success style="font-size:25px"> Continuity And Differentiability L-1 </Success>
### <primary style="font-size:24px"> Geometric interpretation </primary>
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- The ratio $\frac{f(a+h)-f(a)}{h}$ is the slope of the line joining the points $P\equiv(a, f(a))$ and $Q \equiv(a+h, f(a+h))$
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<footer style = "font-size: 18px">Example 2 $\rightarrow$ Differentiability of a function at a point $\rightarrow$ <span style = "color:blue"> Geometric interpretation </span> $\rightarrow$ Geometric interpretation $\rightarrow$ Differentiability of a function at a point </footer>

### <Success style="font-size:25px"> Continuity And Differentiability L-1 </Success>
### <primary style="font-size:24px"> Geometric interpretation </primary>
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- As $h \rightarrow 0$ the point $Q \rightarrow P$, and the secant (ie. the line segment joining $P$ and Q) approaches the tangent line at the point $(a, f(a))$

- So, $\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$ is the slope of the tangent, provided the limit exists.
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<footer style = "font-size: 18px">Differentiability of a function at a point $\rightarrow$ Geometric interpretation $\rightarrow$ <span style = "color:blue"> Geometric interpretation </span> $\rightarrow$ Differentiability of a function at a point $\rightarrow$ Examples </footer>

### <Success style="font-size:25px"> Continuity And Differentiability L-1 </Success>
### <primary style="font-size:24px"> Differentiability of a function at a point </primary>
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- If $f(x)$ is differentiable at $x=a$, then we denote by $f^{\prime}(a)=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$ and $f^{\prime}(a)$
-  is called the "derivative" of $f(x)$ at $x=a$.   
- If the limit of the difference quotient $\frac{f(a+h)-f(a)}{h}$ does not exist, then we say that $f(x)$ is not differentiable at $x=a$.
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<footer style = "font-size: 18px">Geometric interpretation $\rightarrow$ Geometric interpretation $\rightarrow$ <span style = "color:blue"> Differentiability of a function at a point </span> $\rightarrow$ Examples $\rightarrow$ Examples </footer>

### <Success style="font-size:25px"> Continuity And Differentiability L-1 </Success>
### <primary style="font-size:24px"> Examples </primary>
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- (1) $\quad f: \mathbb{R} \rightarrow \mathbb{R}, f(x)=c$, a constant. $\forall x \in \mathbb{R}$.

- For any $a \in \mathbb{R}, \quad f(a+h)-f(a)=c-c=0 \quad\forall h$

- $\therefore \lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}=0$

- $\therefore f$ is differentiable at every a and $f^{\prime}(a)=0$.
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<footer style = "font-size: 18px">Geometric interpretation $\rightarrow$ Differentiability of a function at a point $\rightarrow$ <span style = "color:blue"> Examples </span> $\rightarrow$ Examples $\rightarrow$ Examples </footer>

### <Success style="font-size:25px"> Continuity And Differentiability L-1 </Success>
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- (2) $f(x)=x$
- $f^{\prime}(a) =\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$   
- $=\lim _{h \rightarrow 0} \frac{(a+h)-\alpha}{h}=\lim _{h \rightarrow 0} \frac{k^{\prime}}{k}=1$  
- So, for $f(x)=x, f^{\prime}(x)=1$ for all $x \in \mathbb{R}$

- (3) $f(x)=\{x^2}$

- $\frac{f(x+h)-f(x)}{h}  =\frac{(x+h)^2-x^2}{h}, h \neq 0 $

- $ =\frac{x^2+2 h x+h^2-x^2}{h}$

- $ =\frac{\not h(2 x+h)}{\not h}=2 x+h$

- $ \rightarrow 2 x \text { as } h \rightarrow 0$

- $\therefore f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=2 x \text {. }$

- Notation: $f^{\prime}(x)$ is also denoted by $\frac{d}{d x} f(x)$ is $\frac{d y}{d x}$, where $y=f(x)$.

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<footer style = "font-size: 18px">Differentiability of a function at a point $\rightarrow$ Examples $\rightarrow$ <span style = "color:blue"> Examples </span> $\rightarrow$ Examples $\rightarrow$ Properties </footer>

<h3 id="success-stylefont-size25px-continuity-and-differentiability-l-1-success-1"><Success style="font-size:25px"> Continuity And Differentiability L-1 </Success></h3>
<h3 id="primary-stylefont-size24px-properties-primary"><primary style="font-size:24px"> Properties </primary></h3>
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<p>Property: (1) Let $f(x)$ and $g(x)$ be differentiable at $x=a$. Then $f(x)+g(x)$ is also diff at $x=a$, and</p>
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<p>$\frac{d}{d x}|_{x=a}(f(x)+g(x))$=</p>
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<p>$\frac{d}{d x}|_{x=a} f(x)$</p>
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<p>+$\frac{d}{d x}|_{x=a}g(x)=f^{\prime}(a)+g^{\prime}(a)$</p>
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<footer style = "font-size: 18px">Examples $\rightarrow$ Examples $\rightarrow$ <span style = "color:blue"> Properties </span> $\rightarrow$ Properties $\rightarrow$ Properties </footer>

### <Success style="font-size:25px"> Continuity And Differentiability L-1 </Success>
### <primary style="font-size:24px"> Properties </primary>
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- Proof: Let $u(x)=f(x)+g(x)$

- for $h \neq 0, \frac{u(a+h)-u(a)}{h}=[f(a+h)+g(a+h)]-[f(a)+g(a)]$

- $=\frac{f(a+h)-f(a)}{h}+\frac{g(a+h)-g(a)}{h}$

- $\therefore \lim _{h \rightarrow 0} \frac{u(a+h)-u(a)}{h}= \lim _{h \rightarrow 0} \frac{f(a+h)}{h}-f(a)$

- $+\lim _{h \rightarrow 0} \frac{g(a+h)-g(a)}{h}$

- (by the sum Rule of limits)

- ie. $u^{\prime}(a)=f^{\prime}(a)+g^{\prime}(a)$

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<footer style = "font-size: 18px">Examples $\rightarrow$ Properties $\rightarrow$ <span style = "color:blue"> Properties </span> $\rightarrow$ Properties $\rightarrow$ Example 3 </footer>

<h3 id="success-stylefont-size25px-continuity-and-differentiability-l-1-success-2"><Success style="font-size:25px"> Continuity And Differentiability L-1 </Success></h3>
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<p>(2) If $u(x)=c f(x)$ for some $c \in \mathbb{R}$, & $f^{\prime}(a)$ exists, then $u^{\prime}(a)$ exists and $u^{\prime}(a)=c f^{\prime}(a)$.</p>
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<p>Proof: For $h \neq 0, \frac{u(a+h)-u(a)}{h}=\frac{c f(a+h)-c f(a)}{h}$ $=c\left[\frac{f(a+h)-f(a)}{h}\right]$</p>
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<p>$\rightarrow c f^{\prime}(a)$ as $h \rightarrow 0$.</p>
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<p>$\therefore u^{\prime}(a)=c f^{\prime}(a) \text {. }$</p>
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<p>(3) If $u(x)=c_1 f(x)+c_2 g(x)$ and $f^{\prime}(a), g^{\prime}(a)$ exists then</p>
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<p>$u^{\prime}(a)=c_1 f^{\prime}(a)+c_2 g^{\prime}(a) $.</p>
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<p>Proof: $u^{\prime}(a)=\frac{d}{d x}|_{x=a}(c_1 f(x))$</p>
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<p>$+\frac{d}{d x}|_{x=a}(c_2 g(x))$</p>
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<p>$=c_1 f^{\prime}(a)+c_2 g^{\prime}(a)$</p>
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<p>(by the previous the results).</p>
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<footer style = "font-size: 18px">Properties $\rightarrow$ Properties $\rightarrow$ <span style = "color:blue"> Properties </span> $\rightarrow$ Example 3 $\rightarrow$ Thank you </footer>