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.
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. 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).
\(f(x)=\dfrac{2x+1}{x-2}\) is continuous for all \(x\) except \(\underbrace{x=2}_{not \,in \,the\,domain}\)
A \(x=–8\)
B \(x=12\)
C \(x=3\)
D \(x=–6\)
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 \(x=a\).
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 \(x=–4\).
\(\lim\limits_{x\to a}f(g(x)) =f\left(\lim\limits_{x\to a }g(x)\right)\)= \(f(b)\). 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 \(f(x)=cos\,x\)
\(\therefore\,\lim\limits_{x\to 2}\,cos(x-2)\)
\(=cos\left(\lim\limits_{x\to 2}\,(x-2)\right)\)
\(=cos\,0=1\)
All trigonometric functions are continuous at all points where they are defined i.e they are continuous in their domain.
A \(\dfrac{–1}{5}\)
B 7
C –18
D \(\dfrac{2}{3}\)
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.
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].
Where \(x=\alpha\) is any interior point.
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 \(f(x)=\dfrac{cos\,x}{x–4}\)
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 = \(f(2)\Rightarrow\,\alpha=1\)
A \(\dfrac{3}{2}\)
B \(\dfrac{5}{11}\)
C –10
D \(\dfrac{–11}{7}\)