Solution:

Fix any non zero real number c, we have

` lim_(x->c) f(x) = lim_(x->c) 1/x = 1/c`

Also, since for `c ≠ 0, f (c) = 1/c`, we have ` lim_(x->c) f(x) = f(c)` and hence, f is continuous

at every point in the domain of f. Thus f is a continuous function.

We take this opportunity to explain the concept of infinity. This we do by analysing
the function `f (x) =1/x` near x = 0. To carry out this analysis we follow the usual trick of
finding the value of the function at real numbers close to 0. Essentially we are trying to
find the right hand limit of f at 0. We tabulate this in the following (Table 5.1).

We observe that as x gets closer to 0 from the right, the value of f (x) shoots up
higher. This may be rephrased as: the value of f (x) may be made larger than any given
number by choosing a positive real number very close to 0. In symbols, we write

`lim_(x->0^+) f(x) = + oo`

(to be read as: the right hand limit of f (x) at 0 is plus infinity). We wish to emphasise
that + ∞ is NOT a real number and hence the right hand limit of f at 0 does not exist (as
a real number).
Similarly, the left hand limit of f at 0 may be found. The following table is self
explanatory.

From the Table 5.2, we deduce that the
value of f (x) may be made smaller than any
given number by choosing a negative real
number very close to 0. In symbols,
we write


`lim_(x-> 0^-) f(x) = - oo`

(to be read as: the left hand limit of f (x) at 0 is
minus infinity). Again, we wish to emphasise
that – ∞ is NOT a real number and hence the
left hand limit of f at 0 does not exist (as a real
number). The graph of the reciprocal function
given in Fig 5.3 is a geometric representation
of the above mentioned facts.

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Solution:

The function f is defined at all points of the real line.

Case 1 If c < 1, then f (c) = c + 2. Therefore, `lim_(x->c) f(x) = lim_(x->c) f(x+2) = c+2`

Thus, f is continuous at all real numbers less than 1.


Case 2 If c > 1, then f (c) = c – 2. Therefore,

`lim_(x->c) f(x) = lim_(x->c) (x-2) = c-2 = f(c)`

Thus, f is continuous at all points x > 1.

Case 3 If c = 1, then the left hand limit of f at
x = 1 is

`lim_(x->1^-) f(x) = lim_(x->1^-) (x+2) = 1+2 = 3`


The right hand limit of f at x = 1 is

`lim_(x->1^+) f(x) = lim_(x->1^+) (x-2) = 1-2 = -1`


Since the left and right hand limits of f at x = 1
do not coincide, f is not continuous at x = 1. Hence
x = 1 is the only point of discontinuity of f. The graph of the function is given in Fig 5.4.

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Solution:

As in the previous example we find that f
is continuous at all real numbers x ≠ 1. The left
hand limit of f at x = 1 is

`lim_(x->1^-) f(x) = im_(x-> 1^-) (x+2) = 1+2 =3`

The right hand limit of f at x = 1 is

`lim_(x->1^+) f(x) = lim_(x->1^+) (x-2) = 1-2 = -1`

Since, the left and right hand limits of f at x = 1
do not coincide, f is not continuous at x = 1. Hence
x = 1 is the only point of discontinuity of f. The
graph of the function is given in the Fig 5.5.

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Solution:

Observe that the function is defined at all real numbers except at 0. Domain
of definition of this function is

`D_1 ∪ D_2` where `D_1 = {x ∈ R : x < 0}` and

`D_2 = {x ∈ R : x > 0}`


Case 1 If `c ∈ D_1`, then `lim_(x->c) f(x) = lim_(x->c) (x+2)`

`= c + 2 = f (c) `and hence f is continuous in `D_1`

Case 2 If `c ∈ D_2`, then `lim_(x->c) f(x) = lim_(x->c) (-x+2) `

`= – c + 2 = f (c)` and hence f is continuous in `D_2` .
Since f is continuous at all points in the domain of f,
we deduce that f is continuous. Graph of this
function is given in the Fig 5.6. Note that to graph
this function we need to lift the pen from the plane
of the paper, but we need to do that only for those points where the function is not
defined.

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Solution:

Clearly the function is defined at
every real number. Graph of the function is
given in Fig 5.7. By inspection, it seems prudent
to partition the domain of definition of f into
three disjoint subsets of the real line.


Let `D_1 = {x ∈ R : x < 0}, D_2 = {0}` and

`D_3 = {x ∈ R : x > 0}`

Case 1 At any point in `D_1`, we have `f (x) = x^2` and it is easy to see that it is continuous
there (see Example 2).

Case 2 At any point in `D_3`, we have f (x) = x and it is easy to see that it is continuous
there (see Example 6).

Case 3 Now we analyse the function at x = 0. The value of the function at 0 is f (0) = 0.
The left hand limit of f at 0 is


`lim_(x->0^-) f(x) = lim_(x->0^-) x^2 = 0^2 = 0`

The right hand limit of f at 0 is

`lim_(x->0^+) f(x) = lim_(x->0^+) x= 0`

Thus `lim_(x->0) f(x) = 0 = f(0) ` and hence f is continuous at 0. This means that f is
continuous at every point in its domain and hence, f is a continuous function.

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Solution:

Recall that a function p is a polynomial function if it is defined by
`p(x) = a_0 + a_1 x + ... + a_n x^n` for some natural number n, an ≠ 0 and ai ∈ R. Clearly this
function is defined for every real number. For a fixed real number c, we have

`lim_(x->c) p (x) = p (c)`

By definition, p is continuous at c. Since c is any real number, p is continuous at
every real number and hence p is a continuous function.

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Solution:

First observe that f is defined for all real numbers. Graph of the function is
given in Fig 5.8. From the graph it looks like that f is discontinuous at every integral
point. Below we explore, if this is true.


Case 1 Let c be a real number which is not equal to any integer. It is evident from the
graph that for all real numbers close to c the value of the function is equal to [c]; i.e.,

`lim_(x->c) f(x) = lim_(x->c) [x] = [c]`.

Also `f (c) = [c]` and hence the function is continuous at all real
numbers not equal to integers.


Case 2 Let c be an integer. Then we can find a sufficiently small real number
r > 0 such that [c – r] = c – 1 whereas [c + r] = c.
This, in terms of limits mean that

`lim_(x->c^-) f(x) = c-1 , lim_(x->c^+) f(x) = c`

Since these limits cannot be equal to each other for any c, the function is
discontinuous at every integral point.

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Solution:

Recall that every rational function f is given by

`f(x) = (p (x) )/( q(x)) , q(x) ≠ 0`

where p and q are polynomial functions. The domain of f is all real numbers except
points at which q is zero. Since polynomial functions are continuous (Example 14), f is
continuous by (4) of Theorem 1.

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Solution:

To see this we use the following facts

`lim_(x->0) sin x = 0`

We have not proved it, but is intuitively clear from the graph of sin x near 0.
Now, observe that f (x) = sin x is defined for every real number. Let c be a real
number. Put x = c + h. If x → c we know that h → 0. Therefore

`lim_(x->c) f(x) = lim_(x->c) sin x`

`= lim_(h->0) sin (c+ h )`

`= lim_(h->0) [ sin c cos h + cos c sin h ]`

`= lim_(h->0) [ sin c cos h ] + lim_(h->0) [ cos c sin h ]`

`= sin c + 0 = sin c = f(c)`

Thus `lim_(x->c) f(x) = f(c)` and hence f is a continuous function.

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Solution:

The function f (x) = tan x = `(sin x)/(cos x)` This is defined for all real numbers such
that cos x ≠ 0, i.e., x ≠ (2n +1) `pi/2` . We have just proved that both sine and cosine
functions are continuous. Thus tan x being a quotient of two continuous functions is
continuous wherever it is defined.

An interesting fact is the behaviour of continuous functions with respect to
composition of functions. Recall that if f and g are two real functions, then

`(f o g) (x) = f ( g(x) )`

is defined whenever the range of g is a subset of domain of f. The following theorem
(stated without proof) captures the continuity of composite functions.

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Solution:

Observe that the function is defined for every real number. The function
f may be thought of as a composition g o h of the two functions g and h, where
`g (x) = sin x` and `h (x) = x^2` . Since both g and h are continuous functions, by Theorem 2,
it can be deduced that f is a continuous function.

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Solution:

Define g by `g (x) = 1 – x + | x|` and h by `h (x) = | x|` for all real x. Then

`(h o g) (x) = h (g (x))`

`= h (1– x + | x |)`

`= | 1– x + | x | | = f (x)`

In Example 7, we have seen that h is a continuous function. Hence g being a sum
of a polynomial function and the modulus function is continuous. But then f being a
composite of two continuous functions is continuous

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