
Prove that sine function is continuous at every real number.
Answer
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Hint: We are asked to prove that the sine function is continuous at every real number. First recall the condition necessary for a function to be continuous. Then apply the required limits and check whether the sine function is continuous for every real number.
Complete step-by-step answer:
Let \[f(x) = \sin x\]
We recall the condition to check continuity of function.
Let \[c\] be any real number.
A function \[f(x)\] is continuous at \[x = c\] when it satisfies the condition,
\[L.H.L = R.H.L = f(c)\] (i)
That is when approached from the left hand side and the right hand side we get the same value which means the function is continuous.
Equation (i) can also be written as,
\[L.H.L = R.H.L = f(c)\]
\[ \Rightarrow \mathop {\lim }\limits_{x \to {c^ - }} f(x) = \mathop {\lim }\limits_{x \to {c^ + }} f(x) = f(c)\]
where \[{c^ - }\] can be written as, \[{c^ - } = c - h\] and \[{c^ + }\] can be written as, \[{c^ + } = c + h\] and \[h < < c\]
Now, we take the L.H.L for the function \[f(x) = \sin x\] ,
\[L.H.L = \mathop {\lim }\limits_{x \to {c^ - }} f(x) = \mathop {\lim }\limits_{x \to {c^ - }} \sin x\]
\[ \Rightarrow L.H.L = \mathop {\lim }\limits_{x \to c - h} \sin x\]
\[ \Rightarrow L.H.L = \mathop {\lim }\limits_{h \to 0} \sin \left( {c - h} \right)\]
Applying the trigonometric identity, \[\sin (A - B) = \sin A\cos B - \cos A\sin B\] for the term \[\sin \left( {c - h} \right)\] for the above equation we have,
\[L.H.L = \mathop {\lim }\limits_{h \to 0} \left( {\sin \left( c \right)\cos \left( h \right) - \cos \left( c \right)\sin \left( h \right)} \right)\]
\[ \Rightarrow L.H.L = \sin c\cos 0 - \cos c\sin 0\]
\[ \Rightarrow L.H.L = \sin c.1 - \cos c.0\]
\[ \Rightarrow L.H.L = \sin c\] (i)
Now, we take R.H.L for the function \[f(x) = \sin x\]
\[R.H.L = \mathop {\lim }\limits_{x \to {c^ + }} f(x) = \mathop {\lim }\limits_{x \to {c^ + }} \sin x\]
\[ \Rightarrow R.H.L = \mathop {\lim }\limits_{x \to c + h} \sin x\]
\[ \Rightarrow R.H.L = \mathop {\lim }\limits_{h \to 0} \sin \left( {c + h} \right)\]
Applying the trigonometric identity, \[\sin (A + B) = \sin A\cos B + \cos A\sin B\] for the term \[\sin \left( {c + h} \right)\] for the above equation we have,
\[R.H.L = \mathop {\lim }\limits_{h \to 0} \left( {\sin \left( c \right)\cos \left( h \right) + \cos \left( c \right)\sin \left( h \right)} \right)\]
\[ \Rightarrow R.H.L = \sin c\cos 0 + \cos c\sin 0\]
\[ \Rightarrow R.H.L = \sin c.1 + \cos c.0\]
\[ \Rightarrow R.H.L = \sin c\] (ii)
Comparing equations (i) and (ii) we get,
\[ \Rightarrow L.H.L = R.H.L = \sin c\] (iii)
We have
That is \[x = c\] , \[f(c) = \sin c\]
Equation (iii) becomes,
\[L.H.L = R.H.L = f(c)\]
Therefore, the function sine is continuous for every real number.
Note: For any function, if we are asked to check or prove whether the function is continuous then take the left hand limit and right hand limit on the function and check whether you get the both results equal. Also, remember a function \[f(x)\] is continuous for \[x = a\] only when \[f(a)\] exists and is finite.
Complete step-by-step answer:
Let \[f(x) = \sin x\]
We recall the condition to check continuity of function.
Let \[c\] be any real number.
A function \[f(x)\] is continuous at \[x = c\] when it satisfies the condition,
\[L.H.L = R.H.L = f(c)\] (i)
That is when approached from the left hand side and the right hand side we get the same value which means the function is continuous.
Equation (i) can also be written as,
\[L.H.L = R.H.L = f(c)\]
\[ \Rightarrow \mathop {\lim }\limits_{x \to {c^ - }} f(x) = \mathop {\lim }\limits_{x \to {c^ + }} f(x) = f(c)\]
where \[{c^ - }\] can be written as, \[{c^ - } = c - h\] and \[{c^ + }\] can be written as, \[{c^ + } = c + h\] and \[h < < c\]
Now, we take the L.H.L for the function \[f(x) = \sin x\] ,
\[L.H.L = \mathop {\lim }\limits_{x \to {c^ - }} f(x) = \mathop {\lim }\limits_{x \to {c^ - }} \sin x\]
\[ \Rightarrow L.H.L = \mathop {\lim }\limits_{x \to c - h} \sin x\]
\[ \Rightarrow L.H.L = \mathop {\lim }\limits_{h \to 0} \sin \left( {c - h} \right)\]
Applying the trigonometric identity, \[\sin (A - B) = \sin A\cos B - \cos A\sin B\] for the term \[\sin \left( {c - h} \right)\] for the above equation we have,
\[L.H.L = \mathop {\lim }\limits_{h \to 0} \left( {\sin \left( c \right)\cos \left( h \right) - \cos \left( c \right)\sin \left( h \right)} \right)\]
\[ \Rightarrow L.H.L = \sin c\cos 0 - \cos c\sin 0\]
\[ \Rightarrow L.H.L = \sin c.1 - \cos c.0\]
\[ \Rightarrow L.H.L = \sin c\] (i)
Now, we take R.H.L for the function \[f(x) = \sin x\]
\[R.H.L = \mathop {\lim }\limits_{x \to {c^ + }} f(x) = \mathop {\lim }\limits_{x \to {c^ + }} \sin x\]
\[ \Rightarrow R.H.L = \mathop {\lim }\limits_{x \to c + h} \sin x\]
\[ \Rightarrow R.H.L = \mathop {\lim }\limits_{h \to 0} \sin \left( {c + h} \right)\]
Applying the trigonometric identity, \[\sin (A + B) = \sin A\cos B + \cos A\sin B\] for the term \[\sin \left( {c + h} \right)\] for the above equation we have,
\[R.H.L = \mathop {\lim }\limits_{h \to 0} \left( {\sin \left( c \right)\cos \left( h \right) + \cos \left( c \right)\sin \left( h \right)} \right)\]
\[ \Rightarrow R.H.L = \sin c\cos 0 + \cos c\sin 0\]
\[ \Rightarrow R.H.L = \sin c.1 + \cos c.0\]
\[ \Rightarrow R.H.L = \sin c\] (ii)
Comparing equations (i) and (ii) we get,
\[ \Rightarrow L.H.L = R.H.L = \sin c\] (iii)
We have
That is \[x = c\] , \[f(c) = \sin c\]
Equation (iii) becomes,
\[L.H.L = R.H.L = f(c)\]
Therefore, the function sine is continuous for every real number.
Note: For any function, if we are asked to check or prove whether the function is continuous then take the left hand limit and right hand limit on the function and check whether you get the both results equal. Also, remember a function \[f(x)\] is continuous for \[x = a\] only when \[f(a)\] exists and is finite.
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