Q. 1 - Calculus

Question

As you have already seen, sometimes$\underset{x\to c}{lim} f\left(x\right)$ is equal to f(c), and sometimes it is not. As we will see in Section 1.4, when the limit of a function f as x → c does happen to be equal to the value of f(x) at x = c, we say that the function f is continuous at x = c.

 

(a)   You may have heard the following loose, only partially accurate definition of continuity in a previous class: A function is continuous if you “can draw it without picking up your pencil.” Why does it make sense that this would be related to the definition just presented of continuity in terms of limits? 

(b)   State the limit-definition of continuity with a formal δ–$\in$ statement .

(c)   The function $f\left(x\right)=x^2$ is continuous at every point. Use this fact and the formal definition of continuity to calculate $\underset{x\to 2}{lim} x^2,\underset{x\to 5}{lim} x^2,\text{ and }\underset{x\to -4}{lim} x^2$ .

Step-by-Step Solution

Answer

Ans: 

(a)  If a function is differentiable at a point then it is continuous at the point  

(b)  If for every number ∈>0 there is some number δ>0 such that

         $\left|f\right(x)-L|<\in \text{ whenever }0<|x-a|<\delta$

       There is some number δ>0 such that 

         $\left|f\right(x)-f(a\left)\right|<\in \text{ whenever }0<|x-a|<\delta$

(c)   The function $f\left(x\right)=x^2$ is continuous at every point.

       

Step 1. Given information.

given, 

      As you have already seen, sometimes $\underset{x\to c}{lim} f\left(x\right)$ is equal to f(c), and sometimes it is not. As we will see in Section 1.4, when the limit of a function f as x → c does happen to be equal to the value of f(x) at x = c, we say that the function f is continuous at x = c. 

Step 2. (a) Solution:

Concept Used:-

The left derivative and right derivative of a function f at a point z=c are, equal

    $\begin{array}{r}{f}_{-}^{\mathrm{\prime }}\left(c\right)=\underset{h\to 0^{-}}{lim} \frac{f(c-h)-f\left(c\right)}{-h}\\ f_{+}^{\mathrm{\prime }}\left(c\right)=\underset{h\to 0^{+}}{lim} \frac{f(c+h)-f\left(c\right)}{h}\end{array}$

The derivative of a real-valued function f(z) with respect to x is the function $f\text{'}\left(z\right)$ and is defined as

      $f^{\mathrm{\prime }}\left(z\right)=\underset{h\to 0}{lim} \frac{f(z+h)-f\left(z\right)}{h}$

A function is said to be differentiable if the derivative of the function exists at all points of its domain

The function f is differentiable at z=c  

The derivative of a real-valued function f(z) with respect to x is the function $f\text{'}\left(z\right)$ and is defined as

    $f^{\mathrm{\prime }}\left(z\right)=\underset{h\to 0}{lim} \frac{f(z+h)-f\left(z\right)}{h}$

A function is said to be differentiable if the derivative of the function exists at all points of its domain

The function is differentiable at z=c

That means the existence of the limit

   $f^{\mathrm{\prime }}\left(c\right)=\underset{h\to 0}{lim} \frac{f(c+h)-f\left(c\right)}{h}$ must exist

or

  $f^{\mathrm{\prime }}\left(c\right)=\underset{x\to c}{lim} \frac{f\left(z\right)f\left(c\right)}{z-c}$

Multiplying by $\frac{z-c}{z-c}$

  $\begin{array}{r}f\left(z\right)-f\left(c\right)=\frac{z-c}{z-c}\left\{f\right(z)-f(c\left)\right\} \\ f\left(z\right)-f\left(c\right)=(z-c)\left\{\frac{f\left(z\right)-f\left(c\right)}{z-c}\right\}\end{array}$

taking the limit of both sides 

  $\underset{z\to c}{lim} f\left(z\right)-f\left(c\right)=(z-c)\left\{\underset{x\to c}{lim} \frac{f\left(z\right)-f\left(c\right)}{z-c}\right\}$

The limit of a product is the product of the limits

   $\underset{z\to c}{lim} f\left(z\right)=\underset{z\to c}{lim} f\left(c\right)$

If a function is differentiable at a point then it is continuous at the point 

Step 3. (b) The limit-definition of continuity with a formal &delta;&ndash; &#8712; statement

 Let f(x) be a function defined on an interval that contains x=a, except possibly at x=a

Then

  $\underset{x\to a}{lim} f\left(x\right)=L$

If for every number ∈>0 there is some number δ>0 such that

    $\left|f\right(x)-L|<\in \text{ whenever }0<|x-a|<\delta$

Another definition of limit

Let f(x) be a function defined on an interval that contains x=a

Then f(x) is continuous at x=a

If for every number ∈>0

There is some number δ>0 such that

    $\left|f\right(x)-f(a\left)\right|<\in \text{ whenever }0<|x-a|<\delta$

Step 3. Check continuity:

Continuity at $x\to 2$

LHL: $\underset{x\to 2-}{lim} x^2=4$

RHL:  $\underset{x\to 2+}{lim} x^2=4$

Since LHL=RHL 

Thus the function $\underset{x\to 2}{lim} x^2$ is continuous.  

Continuity at $x\to 5$

LHL:   $\underset{x\to 5^{-}}{lim} x^2=25$

RHL:  $\underset{x\to 5^{+}}{lim} x^2=25$

Since LHL=RHL

Thus the function $\underset{x\to 5}{lim} x^2$ is continuous.  

Continuity at $x\to -4$

LHL   $\underset{x\to -4^{-}}{lim} x^2=16$

RHL   $\underset{x\to -4^{+}}{lim} x^2=16$

Since LHL=RHL

Thus the function $\underset{x\to -4}{lim} x^2$ is continuous.