Question

Solve the given question. Note : ! is 1 actually

Answer 1

✪ᴀɴsᴡᴇʀ✪

• The function is discontinuous at x = 1.

★ᴇxᴘʟᴀɪɴᴀᴛɪᴏɴ★

Here,

$f(x) = \frac{[x^2] - 1} {x^2 - 1} \:\:if \:\:x^2 ≠ 1$

$\:\:\:\:\:\:\:\:\:\:= 0\:\:\:\:\:\:\:\:\:\:if\:\:x^2 = 1$

Now, at x = 1,

Left Hand Limit

LHL $= \lim \limits_{x\to 1^-}f(x)$

$\:\:\:\:\:\:\:\:\:\: = \lim \limits_{x\to 1^-}\frac{[x^2] - 1} {x^2 - 1}$

Since $0 < x^2 < 1, [x^2] = 0$

$\:\:\:\:\:\:\:\:\:\:= \lim \limits_{x\to 1^-}\frac{0 - 1} {x^2 - 1}$

$\:\:\:\:\:\:\:\:\:\: = \lim \limits_{x\to 1^-}\frac{ - 1} {x^2 - 1}$

This limit tends to $+\infty$ and hence, the left hand limit does not exist. Just from this information, we can conclude that $\lim \limits_{x\to 1}f(x)$ does not exist and hence the function is discontinuous at x = 1.

Just for curiosity let us calculate the right hand limit.

Right Hand Limit

RHL $= \lim \limits_{x\to 1^+}f(x)$

$\:\:\:\:\:\:\:\:\:\: = \lim \limits_{x\to 1^+}\frac{[x^2] - 1}{x^2 - 1}$

Since $1 < x^2 < 2, [x^2] = 1$

$\:\:\:\:\:\:\:\:\:\: = \lim \limits_{x\to 1^+}\frac{1 - 1} {x^2 - 1}$

$\:\:\:\:\:\:\:\:\:\: = \lim \limits_{x\to 1^+}\frac{0} {x^2 - 1}$

$\:\:\:\:\:\:\:\:\:\: = 0$

While the numerator is exactly '0', denominator is a number close to '0' but not exactly '0'. Dividing '0' by a non zero number is '0'.