# CBSE Class 11 - Mathematics - Limits and Derivatives

Q1: Define Calculus.

Answer: Calculus is that branch of mathematics that mainly deals with the study of change in the value of a function as the points in the domain change.

👉Note: The chapter "Limits and Derivatives" is an introduction to Calculus.

👉Calculus is a Latin word meaning ‘pebble’. Ancient Romans used stones for counting.

Q2: Who are called pioneers of Calculus (who invented Calculus)?

Answer: Issac Newton (1642 - 1727) and G. W. Leibnitz(1646 - 1717).

Both of them were invented independently around the 17th century.

Q3: What is the meaning of 'x tends to a' or x → a?

Answer: When x tends to a (x  → a), x is nearly close to a but never equals to a.

e.g. x → 3 means the value of x maybe 2.99 or 2.999 or 2.999...9 is very close to 3 but not exactly equal to 3. Similarly, x may be 3.01, 3.001, 3.0001... from the right side and gets closer to 3.

Q4: What do you mean by the left limit of f(x) and right limit of f(x)?

Answer: Consider a function f(x) = x²

Thus at x = 2, f(2) = 4.

When x is less than 2 but increases and gets closer to 2, f(x) also gets closer to 4.

Thus we say x approaches to 2 from left or f(x) approaches 4.

It is called the left-limit of f(x) as x → a is l. If f(x) is arbitrarily close to l when x is in the left-deleted neighbourhood of a and close to it.

Symbolically, $\lim_{x\rightarrow a^-} f(x) = l$ or f(a⁻) = l.

Similarly, when x is greater than 2 and is decreasing and getting closer to 2, f(x) also goes closer to 4.

We can say x approaches 2 from the right, f(x) approaches 4 from the above.

It is called the right limit of f(x) as x → a is l. If f(x) is arbitrarily close to l when x is in the right-deleted neighbourhood of a and close to it.

Symbolically, $\lim_{x\rightarrow a^+} f(x) = l$ or f(a⁺) = l.

Q5: Given function f(x) = x + 5. Find the limit x → 2

Is the function continuous?

Answer: Let us prepare a table when x approaches 2 from left as well as from right.

As the table shows, the value of f(x) at x = 2 is greater than 6.999 and less than 7.001.

⇒ $\lim_{x\rightarrow 2^-} f(x) = 7$ and $\lim_{x\rightarrow 2^+} f(x) = 7$

Hence $\lim_{x\rightarrow 2} f(x) = 7$

Yes, the function is continuous.

Q6: Some rational functions give rise to meaningless conditions like $\frac{0}{0}$. How can we handle such conditions?

Answer: Using limits. It can be solved by cancelling out the factor that is giving 0 by both numerator and denominator

Q7: Find the limit as x → 0 for $f(x) = \frac{5x + x^2}{x}$

Answer: At x = 0, f(x) = $\frac{0}{0}$, which is meaningless.

It implies

$f(x) = \left\{\begin{matrix} 5 + x, \text{if } x \neq 0 \\ \text{is undefined if x = 0} \end{matrix}\right.$

Function f(x) is not continuous at x = 0.

$f(x) = \lim_{x\rightarrow 0} \frac{5x + x^2}{x}$

$f(x) = \lim_{x\rightarrow 0}\frac{x(5 + x)}{x}$

$f(x) = \lim_{x\rightarrow 0} 5 + x = 5$

Thus limit as x approaches 0, f(x) is 5.

Q8: What are three conditions be true then we can say function f is continuous?

Answer: A function f is said to be continuous at x = a if and only if three conditions are all satisfied:

1. f(a) is defined,

2. $\lim_{x\rightarrow a}$ exists

3. f(a) = $\lim_{x\rightarrow a}f(x)$

Q9: Are there any conditions for a function where a limit does not exist? Explain with an example.

Consider f(x) = $\lim_{x\rightarrow 0} (\frac{-1}{x})$

For x = -0.5, f(x) = 2,

x = -0.1, f(x) = 10,

x = -0.01, f(x) = 100

x = -0.001, f(x) = 1000,

x = 0.5, f(x) = -2

x = 0.1, f(x) = -10,

x = 0.01, f(x) = -100

x = 0.001, f(x) = -1000,

Here LHL  RHL.

The values are not approaching a limit,  the limit does not exist.

Q10: Consider the following function
$\begin{cases} x+ 2 & \text{ if } x\neq 1 \\ 0 & \text{ if } x= 1 \end{cases}$

Find $\lim_{x\rightarrow1 } f(x)$ using tabular method.

Answer: Compute values (Left-hand limit and right-hand limit) for f(x) when x nears 1 from left as well as from the right.

For x = 0.9, f(x) = 2.9
x = 0.99, f(x) = 2.99
x = 0.999, f(x) = 2.999
x = 1.001, f(x) = 3.001
x = 1.01, f(x) = 3.01
x = 1.1, f(x) = 3.1

It is clear that:
LHL =  $\lim_{x\rightarrow1^- } f(x) = 3$
RHL =$\lim_{x\rightarrow1^+ } f(x) = 3$

Here LHL = RHL

Thus $\lim_{x\rightarrow1 } f(x) = 3$

Q11: Can we find limits algebraically (without table computation)?

Answer: Yes (We shall see it in the second part).