Chapter 1 Limit, Continuity and Derivative-I

1.1 Exercise 1 (i)

1.1.1 Question 1

Find the domain of the following functions:

• $y=x^2-4x$

It is a polynomial function. Every real number $R$ satisfies the equation. So the domain is $D\left(f\right)=(-\infty ,\infty )$.

• $y=\frac{1}{x+2}$

Here $x$ can be anything except a value that produces a zero in the denominator i.e $x+2\ne 0$ or $x\ne -2$. So the domain is $D\left(f\right)=(-\infty ,-2)\cup (-2,\infty )$.

• $y=\sqrt{3-x}$

Here

$$\begin{array}{c}3-x\ge 0\\ x\le 3\end{array}$$

Therefore, the domain is $D\left(f\right)=(-\infty ,3]$.

• $y=\frac{1}{\sqrt{1-x^2}}$

Here $1-x^2$ should be $>0$ i.e. 

$$\begin{array}{c}1-x^2>0\\ x^2<1\\ -1<x<1\end{array}$$

Also $1-x^2$ can’t be zero.

$$\begin{array}{cc}1-x^2& \ne 0\\ x^2& \ne 1\\ x& \ne \pm 1\end{array}$$

Thus, the domain is $D\left(f\right)=(-1,1)$.

• $y=\sqrt{3-4x+x^2}$

Here

$$\begin{array}{c}3-4x+x^2\ge 0\\ (x-1)(x-3)\ge 0\end{array}$$

Now, when $x$ is $\le 1$ 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 $\ge 3$, LHS is satisfied. Hence the domain of the function is $D\left(f\right)=(-\infty ,1]\cup [3,\infty )$.

• $y=\frac{x}{x-1}$

Here $x-1$ can’t equal zero i.e $x-1\ne 0$ or $x\ne 1$. The domain is thus $D\left(f\right)=R-\left\{1\right\}$.

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 $y=f\left(x\right)$
• 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:

• $y=\frac{1}{x-2}$

Here $x-2$ can’t equal zero i.e $x-2\ne 0$ or $x\ne 2$. The domain is thus $D\left(f\right)=R-\left\{2\right\}$.

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

Figure 1.1: Plotting of $y=\frac{1}{x-2}$

• $\sqrt{x-1}$

Domain: The function must be $x-1\ge 0$. So $x\ge 1$. Domain is thus $D\left(f\right)=[1,\infty )$.

Range: Plotting $y=f\left(x\right)=\sqrt{x}$ shows the graph above $x$-axis. Plotting $f(x-1)=\sqrt{x-1}$ shifts the curve $1$ unit to the right.

The range is thus $R\left(f\right)=[0,\infty )$.

Figure 1.2: Function $y=\sqrt{x-1}$

• $y=\sqrt{25-x^2}$

Domain: For the given function $25-x^2$ should be $\ge 0$.

$$\begin{array}{c}25-x^2\ge 0\\ x^2\le 25\\ -5\le x\le 5\end{array}$$

So the domain is $D\left(f\right)=[-5,5]$.

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

Figure 1.3: $y=\sqrt{25-x^2}$

• $y=\frac{x^2-25}{x-5}$

Domain: Here $x-5\ne 0$, so $x\ne 5$. Domain is thus $D\left(f\right)=R-\left\{5\right\}$.

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

1.1.3 Question 3

Draw the graphs of the following functions:

• $y=\left|x\right|+|x-1|$

When $x<0$,

$$\begin{array}{cc}y& =-x-(x-1)\\ & =-2x+1\end{array}$$

When $0<x<1$,

$$\begin{array}{cc}y& =x-(x-1)\\ & =1\end{array}$$

When $x>1$,

$$\begin{array}{cc}y& =x+x-1\\ & =2x-1\end{array}$$

The graph is thus plotted as below:

Figure 1.4: $y=|x|+|x-1|$

• $y=\left[x\right],x>0$

• $y=\frac{x^2-1}{x-1}$

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

Figure 1.5: $y=\frac{x^2-1}{x-1}$

• $f\left(x\right)=\{\begin{array}{cc}x-& \text{if}x\le 0\\ x^2& \text{if}0<x<2\end{array}$

When $x\le 0$, $f\left(x\right)$ is a straight line with slope $1$ passing through origin.

When $0<x<2$, the $f\left(x\right)$ 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:

• ${lim}_{x\to 0+0}\left|x\right|$

• ${lim}_{x\to 0+0}\frac{1}{x}$

• ${lim}_{x\to 0-0}\frac{1}{x}$

• ${lim}_{x\to 0}\frac{1+2^{1/x}}{3+2^{1/x}}$

• ${lim}_{x\to 0}(2^{1/x}+2^x+\frac{1}{2^x})$

• ${lim}_{x\to \infty }\frac{x}{x+1}$

• ${lim}_{x\to \infty }\frac{3^x-3^{-x}}{3^x+3^{-x}}$

• ${lim}_{x\to \infty }\frac{\sin x}{x}$

• ${lim}_{x\to 0}\frac{{\tan }^{-1}x}{x}$

• ${lim}_{x\to 0}\sin \left(\frac{1}{x}\right)$

1.2.2 Question 2

• A function $f\left(x\right)$ is defined as follows:

$$f\left(x\right)=\{\begin{array}{cc}1-x^2& \text{if}x>0\\ x+1& \text{if}x\le 0\end{array}$$

$--$Find ${lim}_{x\to 0}f\left(x\right)$.

• A function $f$ is defined as follows:

$$f\left(x\right)=\{\begin{array}{cc}{x}^2& \text{when}x<1\\ 2.5& \text{when}x=1\\ x^2+2& \text{when}x>1\end{array}$$

$--$Does ${lim}_{x\to 1}f\left(x\right)$ exist?

• A function $f\left(x\right)$ is defined below:

$$f\left(x\right)=\{\begin{array}{cc}x& \text{if}0<x<1\\ 2-\frac{1}{x}& \text{if}1\le x<2\\ x-\frac{1}{8}{x}^2& \text{if}x>2\end{array}$$

$--$Show the existence of the limits at $x=1$ and $x=2$. Find ${lim}_{x\to 1}f\left(x\right)$ and ${lim}_{x\to 2}f\left(x\right)$ if they exist.

1.2.3 Question 3

Do the following limits exist?

• ${lim}_{x\to 4}\left[x\right]$ where $\left[x\right]$ is the greatest integer function.

• ${lim}_{x\to 2}(x^2+\sqrt{2-x})$

• ${lim}_{x\to 0}\frac{1+e^{\cot x}}{1-e^{\cot x}}$

1.2.4 Question 4

Using $(\epsilon -\delta )$ definition show that:

• ${lim}_{x\to -2}(3x+5)=-1$

• ${lim}_{x\to 3}(2x^2-5x-1)=4$

• ${lim}_{x\to 2}\frac{1}{x}=\frac{1}{2}$

1.3 Additional problems

1.3.1 Find the domain of following functions

• $f\left(x\right)=\frac{\sqrt{x-4}}{x^2-25}$ Ans: $D\left(f\right)=[4,5)\cup (5,\infty )$

• $f\left(x\right)=\frac{\sqrt{x+3}}{\sqrt{x^2-16}}$ Ans: $D\left(f\right)=(4,\infty )$