Question

JEE Mains Maths question is in the image

Answer 1

it has given that $\displaystyle\lim_{x\to 0}\left(\frac{1-x+|x|}{1+[x]-\lambda}\right)=L$

where [.] is the greatest integer function

we have to find L.

when x > 0

|x| = x , [x] = 0

$\displaystyle\lim_{x\to 0^+}\left(\frac{1-x+|x|}{1+[x]-\lambda}\right)=\displaystyle\lim_{x\to 0^+}\frac{1-x+x}{1-\lambda}$

= 1/(1 - λ)

when x < 0

|x| = -x , [x] = -1

$\displaystyle\lim_{x\to 0^-}\left(\frac{1-x+|x|}{1+[x]-\lambda}\right)=\displaystyle\lim_{x\to 0^-}\frac{1-x-x}{1-1-\lambda}$

= 1/λ

for limit is defined

$\displaystyle\lim_{x\to 0^+}f(x)=\displaystyle\lim_{x\to 0^-}f(x)$

⇒1/(1 - λ) = 1/λ

⇒λ = 1/2

so $\displaystyle\lim_{x\to 0}\left(\frac{1-x+|x|}{1+[x]-\lambda}\right)$ = 1/(1 - λ) = 2 = L

Therefore the value of L = 2