If function fx=x-xx,x 0 =1,x=0, then

MHT CET · Maths · Continuity and Differentiability

If function $f (x) = x - \frac{|x|}{x}, x < 0$ $= x + \frac{|x|}{x}, x > 0$ $= 1, x = 0$ , then

A $\underset{x \to 0^{-}}{l i m} f (x)$ does not exist
B $\underset{x \to 0^{+}}{l i m} f (x)$ does not exist
C $f (x)$ is continuous at x = 0
D $\underset{x \to 0^{-}}{l i m} f (x) \neq \underset{x \to 0^{+}}{l i m} f (x)$

Verified Solution

Answer & Solution

Correct Answer

(C) $f (x)$ is continuous at x = 0

Step-by-step Solution

Detailed explanation

Given function,
$ f(x)=x-\frac{|x|}{x}, x<0 $
Now, at $x =0$
$L H L=\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}\left(x-\frac{|x|}{x}\right)$
$\lim _{x \rightarrow 0}\left(x-\frac{(-x)}{x}\right)=\lim _{x \rightarrow 0}(x+1)=1$
$R H L=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0}+\left(x+\frac{|x|}{x}\right)$
$=\lim _{x \rightarrow 0}(x+1)=1 f(0)=1$
$\therefore \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0} f(x)=f(0)=1$
$\Rightarrow f(x)$ is continuous at $x =0$

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From MHT CET

Explore more questions on app

If function fx=x-xx,x<0 =x+xx,x>0 =1,x=0, then

Step-by-step Solution

If function $f (x) = x - \frac{|x|}{x}, x < 0$ $= x + \frac{|x|}{x}, x > 0$ $= 1, x = 0$ , then