Question

If the function \[f(x)=\left\{ {{(\cos x)}^{\dfrac{1}{x}}},x\ne 0 \right\}\]is continuous at x=0, then the value of k is \[f(x)=\]\[\left\{ k,x=0 \right\}\]
A. 8
B. 1
C. -1
D. None of the above

Answer 1

Hint: For this type of function it is given continuous at x equal to zero, that means it is continuous at left hand limit and right hand limit. By finding the limits, left hand limit and right hand limit we get the value of k.

Complete step-by-step solution -
As \[f(x)\]is continuous at \[x=0\],
\[f\left( {{0}^{-}} \right)=f\left( 0 \right)=f\left( {{0}^{+}} \right)\] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
\[f(x)=\left\{ {{(\cos x)}^{\dfrac{1}{x}}},x\ne 0 \right\}\]
\[\underset{x\to 0}{\mathop{\lim }}\,{{(\cos x)}^{\dfrac{1}{x}}}\]
If we substitute x=0 in the following we will get the value of limit as follows,
\[={{1}^{\infty }}\]
\[{{1}^{\infty }}\]is an indeterminate form we cannot find the value of x for this \[{{1}^{\infty }}\]indeterminate form there is a separate procedure
\[\underset{x\to a}{\mathop{\lim }}\,f(x)=1\] and \[\underset{x\to a}{\mathop{\lim }}\,g(x)=\infty \] then the value of limit is
\[\underset{x\to a}{\mathop{\lim }}\,f{{\left( x \right)}^{g\left( x \right)}}={{e}^{\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)\left[ f(x)-1 \right]}}\]
So, in the given problem \[f(x)=\cos x\] and \[g(x)=\dfrac{1}{x}\]. . . . . . . . . . . . . . . . . . . . . . (2)
Substituting the corresponding values in the formula we will get,
\[={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\dfrac{1}{x}\left[ \cos x-1 \right]}}\]. . . . . . . . . . . . . . . . . . . . . .. . (3)
\[={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\dfrac{1}{x}\left[ 2{{\sin }^{2}}\left( \dfrac{x}{2} \right) \right]}}\]
\[={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\dfrac{1}{x}\left[ -2{{\sin }^{2}}\left( \dfrac{x}{2} \right) \right]\times \dfrac{\dfrac{{{x}^{2}}}{4}}{\dfrac{{{x}^{2}}}{4}}}}\]
\[={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\dfrac{1}{x}\left[ -1 \right]\times \dfrac{{{x}^{^{2}}}}{2}}}\]
\[={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\left[ -1 \right]\times \dfrac{x}{2}}}\]
\[={{e}^{0\times -1}}\]
\[=1\]
Therefore, the given function is continuous when\[\underset{x\to 0}{\mathop{\lim }}\,{{(\cos x)}^{\dfrac{1}{x}}}\]=k
We obtained the values as 1 so the value of k=1
K=1
So the correct option is option (B)

Note: In evaluating the limits of form \[{{1}^{\infty }}\].its limit values cannot be calculated directly and can be calculated by the formula given as if \[\underset{x\to a}{\mathop{\lim }}\,f(x)=1\]and \[\underset{x\to a}{\mathop{\lim }}\,g(x)=\infty \]then \[\underset{x\to a}{\mathop{\lim }}\,f{{\left( x \right)}^{g\left( x \right)}}={{e}^{\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)\left[ f(x)-1 \right]}}\]. Note that this formula is only applicable for limits of \[{{1}^{\infty }}\]which is indeterminate form and we know that \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1\].