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).

Illustration Questions

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 \(\underbrace{x=2}_{not \,in \,the\,domain}\) 

Illustration Questions

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 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 \(R-\{n\pi\}\) \((n\in I)\)

Illustration Questions

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.

Illustration Questions

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 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. 

Illustration Questions

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 \(f(x)=cos\,x\)

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

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

\(=cos\,0=1\)

Illustration Questions

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 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.

Illustration Questions

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 = \(f(2)\Rightarrow\,\alpha=1\)

Illustration Questions

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

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