Informative line

Rules For Continuity

Learn rules of continuity, limit of trig and composite function, practice to finding the value of parameter, given the continuity of a piecewise function & continuous in the interval and solve limits with square roots.

Rules of Continuity

Let $$'f'$$ and $$'g'$$ be two functions which are continuous at $$x=a$$, then the following functions will also be continuous at $$x=a$$.

1. $$f+g$$
2. $$f-g$$
3. $$cf$$
4. $$fg$$
5. $$\dfrac{f}{g}$$ if  $$g(a)\ne0$$

• We say that :

1. Sum of two continuous function is also continuous.

Proof: Let $$h(x)=f(x)+g(x)\,\forall \,x$$

and at $$x=a$$  $$f,\,g$$ are say both continuous.

Then,  $$\lim\limits_{x\to a^+}f(x)=\lim\limits_{x\to a^-}f(x)=f(a)$$

and  $$\lim\limits_{x\to a^+}g(x)=\lim\limits_{x\to a^-}g(x)=g(a)$$

$$h(a)=f(a)+g(a)$$

$$\lim\limits_{x\to a^+}h(x)=\lim\limits_{x\to a^+}(f(x)+g(x))=\lim\limits_{x\to a^+}f(x)+\lim\limits_{x\to a^+}g(x)$$

$$\lim\limits_{x\to a^-}h(x)=\lim\limits_{x\to a^-}(f(x)+g(x))=\lim\limits_{x\to a^-}f(x)+\lim\limits_{x\to a^-}g(x)$$

$$\therefore\,\,\lim\limits_{x\to a^+}h(x)=\lim\limits_{x\to a^-}h(x)=h(a)$$

$$\therefore\,\,h$$ is also continuous at $$x=a$$

Similarly,

2.Difference of two continuous functions is continuous.

3.Product of two continuous functions is continuous.

4.Multiplication by a constant will not change the continuous behavior of a function.

5.Ratio of two continuous functions is continuous. (except where denominator is 0).

Let $$'f'$$ and $$'g'$$ be two function which are both continuous at $$x=2$$ and $$f(2)=5,\,g(2)=7$$,then  $$\lim\limits_{x\to 2}(2f(x)+3g(x))$$equals to

A 31

B -28

C 1

D 7

×

$$'f'$$ and $$'g'$$ are continuous at $$x=2$$.

$$2f$$ and $$3g$$ are also continuous at $$x=2$$.

$$\therefore\,\,2f(x)+3g(x)$$ is also continuous at $$x=2$$.

$$\lim\limits_{x\to 2}(2f(x)+3g(x))=2f(2)+3g(2)$$

$$\lim\limits_{x\to 2}(2f(x)+3g(x))$$$$=2×5+3×7$$

$$\lim\limits_{x\to 2}(2f(x)+3g(x))$$=31

Let $$'f'$$ and $$'g'$$ be two function which are both continuous at $$x=2$$ and $$f(2)=5,\,g(2)=7$$,then  $$\lim\limits_{x\to 2}(2f(x)+3g(x))$$equals to

A

31

.

B

-28

C

1

D

7

Option A is Correct

Continuity Behavior of some Common Functions

1. All polynomials are continuous every where i.e in $$(–\infty,\,\infty)$$.
2. All rational functions are continuous whenever they are defined i.e they are continuous in their domain.
• e.g $$f(x)=2x^3+x^2+x-1$$ is continuous everywhere

$$f(x)=\dfrac{2x+1}{x-2}$$ is continuous for all $$x$$ except $$\underbrace{x=2}_{not \,in \,the\,domain}$$

The point of discontinuity of  $$f(x)=\dfrac{2x^3+7x-1}{x-3}$$ is

A $$x=–8$$

B $$x=12$$

C $$x=3$$

D $$x=–6$$

×

As $$f$$ is a rational function so, it will be discontinuous at a point which is not in its domain.

$$\therefore$$ Discontinuous at denominator = 0 values.

$$\Rightarrow$$discontinuous at $$x=3$$.

The point of discontinuity of  $$f(x)=\dfrac{2x^3+7x-1}{x-3}$$ is

A

$$x=–8$$

.

B

$$x=12$$

C

$$x=3$$

D

$$x=–6$$

Option C is Correct

Continuity of Composite Functions

If  $$'g'$$ is continuous at $$x=a$$ and $$'f'$$ is continuous at $$x=g(a)$$, then the composite function $$f(g(x))$$ is continuous at $$x=a$$.

• e.g  $$h(x)=cos\,x^3$$ is continuous at all values of $$x$$, because $$h(x)=f(g(x))$$ where, $$g(x)=x^3$$ and $$f(x)=cos\,x$$, both of which are continuous for all $$x$$.
• Composite function rule extends for composite of more than two function.

Which of the following statement is true ?

A $$h(x)=sin(2x^2+x+1)$$ is continuous at $$x=2$$.

B $$h(x)=cos(x^2+x+3)$$ is discontinuous at $$x=3$$.

C $$h(x)=sin(5x+7)$$ is discontinuous at $$x=5$$.

D $$h(x)=cos(3x^2+5)$$ is discontinuous at $$x=–4$$.

×

Consider option 'a'.

$$h(x)=sin(2x^2+x+1)=f(g(x))$$ where

$$g(x)=2x^2+x+1\,\,$$.

$$f(x)=sin\,x$$

Both of which are continuous for all values of $$x$$

$$\therefore\,h(x)$$ is continuous $$\forall\; x$$

$$\Rightarrow$$ continuous at $$x=2$$

All other statements are false as function $$h(x)$$ is continuous at indicated values.

Which of the following statement is true ?

A

$$h(x)=sin(2x^2+x+1)$$ is continuous at $$x=2$$.

.

B

$$h(x)=cos(x^2+x+3)$$ is discontinuous at $$x=3$$.

C

$$h(x)=sin(5x+7)$$ is discontinuous at $$x=5$$.

D

$$h(x)=cos(3x^2+5)$$ is discontinuous at $$x=–4$$.

Option A is Correct

Limit of a Composite Function

$$\lim\limits_{x\to a}f(g(x)) =f\left(\lim\limits_{x\to a }g(x)\right)$$$$f(b)$$. If

(1)  $$\lim\limits_{x\to a}\,g(x)=b$$

(2)  $$'f'$$is continuous at $$x=b$$

e.g

Consider,  $$\lim\limits_{x\to a}\,cos(x-2)$$

which is of the form  $$\lim\limits_{x\to a}\,f(g(x))$$

where, $$g(x)=x-2$$ and $$f(x)=cos\,x$$

$$\therefore\,\lim\limits_{x\to 2}\,cos(x-2)$$

$$=cos\left(\lim\limits_{x\to 2}\,(x-2)\right)$$

$$=cos\,0=1$$

The value of  $$\lim\limits_{x\to 0}\,cos(x^2+x)$$ is

A 2

B 1

C 17

D –10

×

$$\lim\limits_{x\to 0}\,cos(x^2+x)=\underbrace {cos\left(\lim\limits_{x\to 0}\,x^2+x\right)}_{Limit\, of\, composite\, function \,rule}$$

$$cos\left(\lim\limits_{x\to 0}\,x^2+x\right)$$

$$=\underbrace {cos(0)}_{Polynomial \,rule\, of\, continuity}=1$$

The value of  $$\lim\limits_{x\to 0}\,cos(x^2+x)$$ is

A

2

.

B

1

C

17

D

–10

Option B is Correct

Continuity of Trigonometric Functions

All trigonometric functions are continuous at all points where they are defined i.e they are continuous in their domain.

• $$sin\,x,\,cos\,x$$ are continuous in R
• $$tan\,x,\,sec\,x$$ are continuous in $$R-\left\{(2n+1)\dfrac{\pi}{2}\right\}$$$$(n\in I)$$
• $$cosec\,x,\,cot\,x$$ are continuous in $$R-\{n\pi\}$$ $$(n\in I)$$

The value of $$\lim\limits_{x\to \pi}\,\dfrac{cos\,x}{5+sin\,x}$$ is.

A $$\dfrac{–1}{5}$$

B 7

C –18

D $$\dfrac{2}{3}$$

×

$$f(x)=\dfrac{cos\,x}{5+sin\,x}$$ is continuous for all $$x$$.

$$cos\,x$$ and  $$5+sin\,x$$ both are continuous at $$x=\pi$$

$$f$$ is continuous by $$\dfrac{f}{g}$$ rule.

$$\therefore\,\lim\limits_{x\to \pi}\,\dfrac{cos\,x}{5+sin\,x}=f(\pi)$$ by continuity.

$$\dfrac{cos\pi}{5+sin\pi}=\dfrac{–1}{5+0}=\dfrac{–1}{5}$$

The value of $$\lim\limits_{x\to \pi}\,\dfrac{cos\,x}{5+sin\,x}$$ is.

A

$$\dfrac{–1}{5}$$

.

B

7

C

–18

D

$$\dfrac{2}{3}$$

Option A is Correct

Continuity of Root Functions

The root function $$f(x)= \sqrt[n]{x}=x^{1/n}$$ is continuous for all values of  $$x$$ where it is defined or continuous in its domain.

• Consider the graph of $$f(x)=x^{1/n}$$.
• For the case, when $$n$$ is even, the graph is similar to  $$f(x)=\sqrt x$$ as shown in figure.

• For the case, when $$n$$ is odd the graph is the graph is similar to $$f(x)=\sqrt[3]{x}$$  as shown in the graph.

• In both cases $$f(x)=x^{{1}/{n}}$$ is continuous in domain.

The value of  $$\lim\limits_{x\to 4}\dfrac{\sqrt x+1}{x+5}$$ is

A $$\dfrac{1}{3}$$

B 7

C –2

D 8

×

$$f(x)=\sqrt x$$ is continuous at $$x=4$$

$$\sqrt x+1$$ is continuous at $$x=4$$ (using f+g rule).

$$\Rightarrow\,\dfrac{\sqrt x+1}{x+5}$$ is continuous at $$x=4$$ (using ratio rule)

$$\therefore\,\lim\limits_{x\to 4}\dfrac{\sqrt x+1}{x+5}=f(4)$$

$$=\dfrac{\sqrt 4+1}{4+5}=\dfrac{3}{9}=\dfrac{1}{3}$$

The value of  $$\lim\limits_{x\to 4}\dfrac{\sqrt x+1}{x+5}$$ is

A

$$\dfrac{1}{3}$$

.

B

7

C

–2

D

8

Option A is Correct

Continuity in an Interval

A function $$'f'$$ is said to be continuous in an interval

[a, b] if it is continuous for all values of  $$x$$ in [a, b].

• At $$x=a$$, continuity means $$\lim\limits_{x\to a^+}\,f(x)=f(a)$$ (continuity from right).
• At $$x=b$$, continuity means $$\lim\limits_{x\to b^-}\,f(x)=f(b)$$ (continuity from left).
• At all the interior points the definition of continuity remain the same i.e R.H.L=L.H.L = $$f(\alpha)$$

Where $$x=\alpha$$ is any interior point.

• Even a single point of discontinuity in any interval will make the function discontinuous in the entire interval.

Which of the following function is continuous in the interval [–5, 7] ?

A $$f(x)=\dfrac{sin\,x}{x–8}$$

B $$f(x)=\dfrac{cos\,x}{1–x}$$

C $$f(x)=\dfrac{sin\,x}{x+2}$$

D $$f(x)=\dfrac{cos\,x}{x–4}$$

×

By the ratio rule of continuity, the functions in the options are discontinuous at $$x=8,\,x=1,\,x=–2$$ and $$x=4$$ respectively (put denominator = 0).

All these values lie in [–5, 7] except $$x=8$$.

$$\therefore\,f(x)=\dfrac{sin\,x}{x–8}$$ is continuous in [–5, 7].

It is continuous in any interval not containing $$x=8$$.

Which of the following function is continuous in the interval [–5, 7] ?

A

$$f(x)=\dfrac{sin\,x}{x–8}$$

.

B

$$f(x)=\dfrac{cos\,x}{1–x}$$

C

$$f(x)=\dfrac{sin\,x}{x+2}$$

D

$$f(x)=\dfrac{cos\,x}{x–4}$$

Option A is Correct

Finding the Value of Parameter when the Continuity of a Piecewise Function is given

Suppose $$'f'$$ is a piecewise defined function.

$$f(x)= \begin{cases} 2x+1 & if & x<2\\ x + 3 & if & x\geq0 \end{cases}$$

then to test continuity at $$f$$ at $$x=2$$

R.H.L = $$\lim\limits_{x\to 2^+}f(x)=\lim\limits_{x\to 2^+}(x+3)=5\;\;\;\;\;\;(2^+>2)$$

L.H.L =  $$\lim\limits_{x\to 2^-}f(x)=\lim\limits_{x\to 2^-}(2x+1)=5\,\;\;\;\;(2^-<2)$$

$$f(2)=2+3=5$$

$$\therefore$$ $$f$$ is continuous at $$x=2$$

Now sometimes it is asked after  giving continuity ,what is the value of certain parameter.

Consider

$$f(x)= \begin{cases} 2x+\alpha & if & x<2\\ x + 3 & if & x\geq2 \end{cases}$$  is continuous at $$x=2$$

then R.H.L = L.H.L = $$f(2)\Rightarrow\,\alpha=1$$

Let $$f(x)= \begin{cases} 2+cx & if & x\leq2\\ x^2+x–1 & if & x>2 \end{cases}$$ The value of 'c' which will make $$'f'$$ continuous in $$(– \infty,\,\infty)$$ is

A $$\dfrac{3}{2}$$

B $$\dfrac{5}{11}$$

C –10

D $$\dfrac{–11}{7}$$

×

$$'f'$$ is a continuous function for all values of $$x$$ except perhaps at $$x=2$$, (being a polynomial on either side of $$x=2$$)

For continuity at $$x=2$$ $$\to$$ R.H.L = L.H.L = $$f(2)$$.

R.H.L = $$\lim\limits_{x\to 2^+}f(x)=\lim\limits_{x\to 2^+}(x^2+x–1)=4+2–1=5$$

L.H.L = $$\lim\limits_{x\to 2^-}f(x)=\lim\limits_{x\to 2^-}(2+cx)=2+2c$$

$$f(2)=2+2c\,\Rightarrow\,5=2+2c\,\Rightarrow\,c=\dfrac{3}{2}$$

Let $$f(x)= \begin{cases} 2+cx & if & x\leq2\\ x^2+x–1 & if & x>2 \end{cases}$$ The value of 'c' which will make $$'f'$$ continuous in $$(– \infty,\,\infty)$$ is

A

$$\dfrac{3}{2}$$

.

B

$$\dfrac{5}{11}$$

C

–10

D

$$\dfrac{–11}{7}$$

Option A is Correct