JEE MAIN - Mathematics (2019 - 9th April Evening Slot - No. 15)
If $$f(x) = [x] - \left[ {{x \over 4}} \right]$$ ,x $$ \in $$
4
, where [x] denotes the
greatest integer function, then
Both $$\mathop {\lim }\limits_{x \to 4 - } f(x)$$ and $$\mathop {\lim }\limits_{x \to 4 + } f(x)$$ exist but are not
equal
f is continuous at x = 4
$$\mathop {\lim }\limits_{x \to 4 + } f(x)$$ exists but $$\mathop {\lim }\limits_{x \to 4 - } f(x)$$ does not exist
$$\mathop {\lim }\limits_{x \to 4 - } f(x)$$ exists but $$\mathop {\lim }\limits_{x \to 4 + } f(x)$$ does not exist
Explanation
$$f(x) = [x] - \left[ {{x \over 4}} \right]$$
Here check continuty at x = 4
LHL = $$\mathop {\lim }\limits_{x \to {4^ - }} f\left( x \right)$$
= $$\mathop {\lim }\limits_{x \to {4^ - }} \left[ x \right] - \left[ {{x \over 4}} \right]$$
= $$\mathop {\lim }\limits_{h \to 0} \left[ {4 - h} \right] - \left[ {{{4 - h} \over 4}} \right]$$
= 3 - 0 = 3
h $$ \to $$ 0 means h > 0 (very small positive value)
So 4 - h = 3.something (means less than 4)
$$ \therefore $$ [4 - h] = [3.something] = 3
and $$\left[ {{{4 - h} \over 4}} \right]$$ = [0.something] = 0
RHL = $$\mathop {\lim }\limits_{x \to {4^ + }} f\left( x \right)$$
= $$\mathop {\lim }\limits_{x \to {4^ + }} \left[ x \right] - \left[ {{4 \over x}} \right]$$
= $$\mathop {\lim }\limits_{h \to 0} \left[ {4 + h} \right] - \left[ {{{4 + h} \over 4}} \right]$$
= 4 - 1 = 3
Here [4 + h] = [4.something] = 4
and $$\left[ {{{4 + h} \over 4}} \right]$$ = [1.something] = 1
And f(4) = $$\left[ 4 \right] - \left[ {{4 \over 4}} \right]$$ = 4 - 1 = 3
As LHL = RHL = f(4)
$$ \therefore $$ f is continuous at x = 4.
Here check continuty at x = 4
LHL = $$\mathop {\lim }\limits_{x \to {4^ - }} f\left( x \right)$$
= $$\mathop {\lim }\limits_{x \to {4^ - }} \left[ x \right] - \left[ {{x \over 4}} \right]$$
= $$\mathop {\lim }\limits_{h \to 0} \left[ {4 - h} \right] - \left[ {{{4 - h} \over 4}} \right]$$
= 3 - 0 = 3
h $$ \to $$ 0 means h > 0 (very small positive value)
So 4 - h = 3.something (means less than 4)
$$ \therefore $$ [4 - h] = [3.something] = 3
and $$\left[ {{{4 - h} \over 4}} \right]$$ = [0.something] = 0
RHL = $$\mathop {\lim }\limits_{x \to {4^ + }} f\left( x \right)$$
= $$\mathop {\lim }\limits_{x \to {4^ + }} \left[ x \right] - \left[ {{4 \over x}} \right]$$
= $$\mathop {\lim }\limits_{h \to 0} \left[ {4 + h} \right] - \left[ {{{4 + h} \over 4}} \right]$$
= 4 - 1 = 3
Here [4 + h] = [4.something] = 4
and $$\left[ {{{4 + h} \over 4}} \right]$$ = [1.something] = 1
And f(4) = $$\left[ 4 \right] - \left[ {{4 \over 4}} \right]$$ = 4 - 1 = 3
As LHL = RHL = f(4)
$$ \therefore $$ f is continuous at x = 4.
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