Chapter 1 Limit, Continuity and Derivative-I

1.1 Exercise 1 (i)

1.1.1 Question 1

Find the domain of the following functions:

  • y=x24x

It is a polynomial function. Every real number R satisfies the equation. So the domain is D(f)=(,).

  • y=1x+2

Here x can be anything except a value that produces a zero in the denominator i.e x+20 or x2. So the domain is D(f)=(,2)(2,).

  • y=3x

Here

3x0x3

Therefore, the domain is D(f)=(,3].

  • y=11x2

Here 1x2 should be >0 i.e. 

1x2>0x2<11<x<1

Also 1x2 can’t be zero.

1x20x21x±1

Thus, the domain is D(f)=(1,1).

  • y=34x+x2

Here

34x+x20(x1)(x3)0

Now, when x is 1 in the number line, LHS is satisfied.

When x lies between (1,3), LHS becomes negative and won’t satisfy the equation.

When x is 3, LHS is satisfied. Hence the domain of the function is D(f)=(,1][3,).

  • y=xx1

Here x1 can’t equal zero i.e x10 or x1. The domain is thus D(f)=R{1}.

Finding range of a function

A. Method 1

  • Plot the graph (use all techniques of shifting, stretching, compression)
  • Range will be the value along y-axis

B. Method 2

  • Put y=f(x)
  • Express x as a function of y
  • Find possible values for y (just like domain)
  • Eliminate values by looking at the definition to write the final range

1.1.2 Question 2

Find the domain and the range of the following functions:

  • y=1x2

Here x2 can’t equal zero i.e x20 or x2. The domain is thus D(f)=R{2}.

Plotting shows that the range of the function (the values along y-axis) is (,0)(0,).

Plotting of $y=\frac{1}{x-2}$

Figure 1.1: Plotting of y=1x2

  • x1

Domain: The function must be x10. So x1. Domain is thus D(f)=[1,).

Range: Plotting y=f(x)=x shows the graph above x-axis. Plotting f(x1)=x1 shifts the curve 1 unit to the right.

The range is thus R(f)=[0,).

Function $y=\sqrt{x-1}$

Figure 1.2: Function y=x1

  • y=25x2

Domain: For the given function 25x2 should be 0.

25x20x2255x5

So the domain is D(f)=[5,5].

Range: Plotting shows the function to be half circle above x-axis. The range is thus values along the y-axis i.e R(f)=[0,5].

$y=\sqrt{25-x^2}$

Figure 1.3: y=25x2

  • y=x225x5

Domain: Here x50, so x5. Domain is thus D(f)=R{5}.

Range: Plotting the equation gets a line y=x+5 except a hole at point (5,10). Range is therefore R(f)=(,10)(10,) or R{10}.

1.1.3 Question 3

Draw the graphs of the following functions:

  • y=|x|+|x1|

When x<0,

y=x(x1)=2x+1

When 0<x<1,

y=x(x1)=1

When x>1,

y=x+x1=2x1

The graph is thus plotted as below:

$y=|x| + |x-1|$

Figure 1.4: y=|x|+|x1|

  • y=[x],x>0

  • y=x21x1

For x1, y=x+1 which is a straight line not passing through the point (1,2) i.e hole at this point. The graph is given below:

$y=\frac{x^2-1}{x-1}$

Figure 1.5: y=x21x1

  • f(x)={xif x0x2if 0<x<2

When x0, f(x) is a straight line with slope 1 passing through origin.

When 0<x<2, the f(x) is a parabola with concavity upwards. The graph is shown below:

1.2 Exercise 1(ii)

1.2.1 Question 1

Find the limits of the following:

  • limx0+0|x|

  • limx0+01x

  • limx001x

  • limx01+21/x3+21/x

  • limx0(21/x+2x+12x)

  • limxxx+1

  • limx3x3x3x+3x

  • limxsinxx

  • limx0tan1xx

  • limx0sin(1x)

1.2.2 Question 2

  • A function f(x) is defined as follows:

f(x)={1x2if x>0x+1if x0

Find limx0f(x).

  • A function f is defined as follows:

f(x)={x2when x<12.5when x=1x2+2when x>1

Does limx1f(x) exist?

  • A function f(x) is defined below:

f(x)={xif 0<x<121xif 1x<2x18x2if x>2

Show the existence of the limits at x=1 and x=2. Find limx1f(x) and limx2f(x) if they exist.

1.2.3 Question 3

Do the following limits exist?

  • limx4[x] where [x] is the greatest integer function.

  • limx2(x2+2x)

  • limx01+ecotx1ecotx

1.2.4 Question 4

Using (εδ) definition show that:

  • limx2(3x+5)=1

  • limx3(2x25x1)=4

  • limx21x=12

1.3 Additional problems

1.3.1 Find the domain of following functions

  • f(x)=x4x225 Ans: D(f)=[4,5)(5,)

  • f(x)=x+3x216 Ans: D(f)=(4,)