# If $f(x) = \dfrac{{{e^{\dfrac{1}{x}}} - 1}}{{{e^{\dfrac{1}{x}}} + 1}},x \ne 0{\text{ and f}}\left( 0 \right) = 0$, then $f(x)$ is

(a) Continuous at $0$

(b) Right continuous at $0$

(c) Discontinuous at $0$

(d) Left continuous at $0$

Last updated date: 17th Mar 2023

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Answer

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Hint- Calculate left hand and right hand limit at the required point where continuity is asked.

We have to comment upon the continuity of $f(x) = \dfrac{{{e^{\dfrac{1}{x}}} - 1}}{{{e^{\dfrac{1}{x}}} + 1}},x \ne 0{\text{ }}$at x=0

Let’s calculate the left hand side limit for this $f(x)$

$ \Rightarrow f{(x)_{x \to {0^ - }}} = f{(0 - h)_{h \to 0}} = f{( - h)_{h \to 0}}$

So $f{(x)_{x \to {0^ - }}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{{e^{\dfrac{{ - 1}}{h}}} - 1}}{{{e^{\dfrac{{ - 1}}{h}}} + 1}}$

We can write this down as

$f{(x)_{x \to {0^ - }}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{\dfrac{1}{{{e^{\dfrac{1}{h}}}}} - 1}}{{\dfrac{1}{{{e^{\dfrac{1}{h}}}}} + 1}}$

Now let’s substitute 0 in place of h we get

$f{(x)_{x \to {0^ - }}} = \dfrac{{\dfrac{1}{{{e^{\dfrac{1}{0}}}}} - 1}}{{\dfrac{1}{{{e^{\dfrac{1}{0}}}}} + 1}}$

Now we know that $\dfrac{1}{0} = \infty {\text{ and }}{{\text{e}}^{\dfrac{1}{0}}} = {e^\infty } = \infty $

Putting it above we get

$ \Rightarrow f{(x)_{x \to {0^ - }}} = \dfrac{{\dfrac{1}{\infty } - 1}}{{\dfrac{1}{\infty } + 1}} = \dfrac{{0 - 1}}{{0 + 1}} = - 1$

So the left hand limit is -1, now let’s compute the Right side limit

$ \Rightarrow f{(x)_{x \to {0^ + }}} = f{(0 + h)_{h \to 0}} = f{(h)_{h \to 0}}$

$f{(x)_{x \to {0^ + }}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{{e^{\dfrac{1}{h}}} - 1}}{{{e^{\dfrac{1}{h}}} + 1}}$

Now we will be taking ${e^{\dfrac{1}{h}}}$ common from the denominator as well as from the numerator

$ \Rightarrow f{(x)_{x \to {0^ + }}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{1 - \dfrac{1}{{{e^{\dfrac{1}{h}}}}}}}{{1 + \dfrac{1}{{{e^{\dfrac{1}{h}}}}}}}$

Now let’s substitute 0 in place of h we get

$ \Rightarrow f{(x)_{x \to {0^ + }}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{1 - \dfrac{1}{{{e^{\dfrac{1}{0}}}}}}}{{1 + \dfrac{1}{{{e^{\dfrac{1}{0}}}}}}}$

Now we know that $\dfrac{1}{0} = \infty {\text{ and }}{{\text{e}}^{\dfrac{1}{0}}} = {e^\infty } = \infty $ putting it above we get

$ \Rightarrow f{(x)_{x \to {0^ + }}} = \dfrac{{1 - \dfrac{1}{\infty }}}{{1 + \dfrac{1}{\infty }}} = \dfrac{{1 - 0}}{{1 + 0}} = 1$

Now clearly the left hand limit at $x = 0$ is not equal to the right hand limit at $x = 0$. Hence given $f(x)$ is discontinuous at $x = 0$

Hence option (c) is correct

Note- Whenever we are told to comment upon the continuity of a given function at a specific point, approach the point first from left side and then from right side if both the limits are equal and it is equal to the value of the function at that point then the function is continuous at that point.

We have to comment upon the continuity of $f(x) = \dfrac{{{e^{\dfrac{1}{x}}} - 1}}{{{e^{\dfrac{1}{x}}} + 1}},x \ne 0{\text{ }}$at x=0

Let’s calculate the left hand side limit for this $f(x)$

$ \Rightarrow f{(x)_{x \to {0^ - }}} = f{(0 - h)_{h \to 0}} = f{( - h)_{h \to 0}}$

So $f{(x)_{x \to {0^ - }}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{{e^{\dfrac{{ - 1}}{h}}} - 1}}{{{e^{\dfrac{{ - 1}}{h}}} + 1}}$

We can write this down as

$f{(x)_{x \to {0^ - }}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{\dfrac{1}{{{e^{\dfrac{1}{h}}}}} - 1}}{{\dfrac{1}{{{e^{\dfrac{1}{h}}}}} + 1}}$

Now let’s substitute 0 in place of h we get

$f{(x)_{x \to {0^ - }}} = \dfrac{{\dfrac{1}{{{e^{\dfrac{1}{0}}}}} - 1}}{{\dfrac{1}{{{e^{\dfrac{1}{0}}}}} + 1}}$

Now we know that $\dfrac{1}{0} = \infty {\text{ and }}{{\text{e}}^{\dfrac{1}{0}}} = {e^\infty } = \infty $

Putting it above we get

$ \Rightarrow f{(x)_{x \to {0^ - }}} = \dfrac{{\dfrac{1}{\infty } - 1}}{{\dfrac{1}{\infty } + 1}} = \dfrac{{0 - 1}}{{0 + 1}} = - 1$

So the left hand limit is -1, now let’s compute the Right side limit

$ \Rightarrow f{(x)_{x \to {0^ + }}} = f{(0 + h)_{h \to 0}} = f{(h)_{h \to 0}}$

$f{(x)_{x \to {0^ + }}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{{e^{\dfrac{1}{h}}} - 1}}{{{e^{\dfrac{1}{h}}} + 1}}$

Now we will be taking ${e^{\dfrac{1}{h}}}$ common from the denominator as well as from the numerator

$ \Rightarrow f{(x)_{x \to {0^ + }}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{1 - \dfrac{1}{{{e^{\dfrac{1}{h}}}}}}}{{1 + \dfrac{1}{{{e^{\dfrac{1}{h}}}}}}}$

Now let’s substitute 0 in place of h we get

$ \Rightarrow f{(x)_{x \to {0^ + }}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{1 - \dfrac{1}{{{e^{\dfrac{1}{0}}}}}}}{{1 + \dfrac{1}{{{e^{\dfrac{1}{0}}}}}}}$

Now we know that $\dfrac{1}{0} = \infty {\text{ and }}{{\text{e}}^{\dfrac{1}{0}}} = {e^\infty } = \infty $ putting it above we get

$ \Rightarrow f{(x)_{x \to {0^ + }}} = \dfrac{{1 - \dfrac{1}{\infty }}}{{1 + \dfrac{1}{\infty }}} = \dfrac{{1 - 0}}{{1 + 0}} = 1$

Now clearly the left hand limit at $x = 0$ is not equal to the right hand limit at $x = 0$. Hence given $f(x)$ is discontinuous at $x = 0$

Hence option (c) is correct

Note- Whenever we are told to comment upon the continuity of a given function at a specific point, approach the point first from left side and then from right side if both the limits are equal and it is equal to the value of the function at that point then the function is continuous at that point.

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