
If the function f(x) $ = \left\{\begin{array}
\;
\dfrac{\log x - \log 4}{x - 4}, \text{ for } x \ne 4 \\
\dfrac{1}{4}, \text{ for } x = 4 \\
\end{array} \right.
$
Find whether the function is continuous at x=4.
Answer
585.3k+ views
Hint: We know that for a particular function to be continuous at a point it’s left limit , right limit and value at that particular point should be equal. So, in order to check whether those are equal or not we have to find those values. Check if the limit values are in the form of $\dfrac{0}{0}$ , if those are in the form of $\dfrac{0}{0}$then use L’Hospital’s rule.
Complete step-by-step answer:
First let us find the left hand limit of the given function at x = 4
That is $\mathop {\lim }\limits_{x \to 4 - } \dfrac{{\log x - \log 4}}{{x - 4}}$
This is in the form of $\dfrac{0}{0}$ when x is placed as 0.
So, L'Hospital's Rule tells us that if we have an indeterminate form 0/0 or ∞/∞ all we need to do is differentiate the numerator and differentiate the denominator and then take the limit.
This says we can use L'Hospital's Rule to sort out the value of this function’s left hand limit.
Which implies the left hand limit will be
$\mathop {\lim }\limits_{x \to 4 - } \dfrac{{d(\log x) - d(\log 4)}}{{d(x - 4)}}\, = \,\mathop {\lim }\limits_{x \to 4 - } \dfrac{{(\dfrac{1}{x})}}{1}\, = \,\dfrac{1}{4}$;
Now let us find the right hand limit of the given function at x = 4
That is $\mathop {\lim }\limits_{x \to 4 + } \dfrac{{\log x - \log 4}}{{x - 4}}$
This is in the form of $\dfrac{0}{0}$ when x is placed as 0.
This says we can use L'Hospital's Rule to sort out the value of this function’s right hand limit also.
$\mathop {\lim }\limits_{x \to 4 + } \dfrac{{d(\log x) - d(\log 4)}}{{d(x - 4)}}\, = \,\mathop {\lim }\limits_{x \to 4 + } \dfrac{{(\dfrac{1}{x})}}{1}\, = \,\dfrac{1}{4}$;
It is given that the value of the given function is ¼ at x=4;
Which implies The left hand limit =Right hand limit =function value at x=4;
So, the given function is continuous at x=4.
Note: Read the properties of limits. Read the properties of derivatives. And also check the conditions of continuity and differentiability. Solve twice while calculating the value of limits.Read the concepts of L’Hospital’s rule and where it should be applied.
Complete step-by-step answer:
First let us find the left hand limit of the given function at x = 4
That is $\mathop {\lim }\limits_{x \to 4 - } \dfrac{{\log x - \log 4}}{{x - 4}}$
This is in the form of $\dfrac{0}{0}$ when x is placed as 0.
So, L'Hospital's Rule tells us that if we have an indeterminate form 0/0 or ∞/∞ all we need to do is differentiate the numerator and differentiate the denominator and then take the limit.
This says we can use L'Hospital's Rule to sort out the value of this function’s left hand limit.
Which implies the left hand limit will be
$\mathop {\lim }\limits_{x \to 4 - } \dfrac{{d(\log x) - d(\log 4)}}{{d(x - 4)}}\, = \,\mathop {\lim }\limits_{x \to 4 - } \dfrac{{(\dfrac{1}{x})}}{1}\, = \,\dfrac{1}{4}$;
Now let us find the right hand limit of the given function at x = 4
That is $\mathop {\lim }\limits_{x \to 4 + } \dfrac{{\log x - \log 4}}{{x - 4}}$
This is in the form of $\dfrac{0}{0}$ when x is placed as 0.
This says we can use L'Hospital's Rule to sort out the value of this function’s right hand limit also.
$\mathop {\lim }\limits_{x \to 4 + } \dfrac{{d(\log x) - d(\log 4)}}{{d(x - 4)}}\, = \,\mathop {\lim }\limits_{x \to 4 + } \dfrac{{(\dfrac{1}{x})}}{1}\, = \,\dfrac{1}{4}$;
It is given that the value of the given function is ¼ at x=4;
Which implies The left hand limit =Right hand limit =function value at x=4;
So, the given function is continuous at x=4.
Note: Read the properties of limits. Read the properties of derivatives. And also check the conditions of continuity and differentiability. Solve twice while calculating the value of limits.Read the concepts of L’Hospital’s rule and where it should be applied.
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