Limits, Continuity & Differentiability
Limit of composite functions with floor and signum
nta_pyq_2023_jan
Grade None
Question:
Let f, g and h be the real valued functions defined on \mathbb{R} as f(x) = \begin{cases} \frac{x}{|x|}, & x \neq 0 \\ 1, & x = 0 \end{cases}, g(x) = \begin{cases} \frac{\sin(x+1)}{(x+1)}, & x \neq -1 \\ 1, & x = -1 \end{cases} and h(x) = 2[x] - f(x), where [x] is the greatest integer \leq x. Then the value of \lim_{x \to 1} g(h(x-1)) is
Step-by-Step Solution
Key Concept: Compute h(x-1) from left and right as x \to 1, using f(x) = \pm 1 and floor function. Then apply g to the result.
LHL: h((1-k)-1) = h(-k) = 2(-1)-(-1) = -1, g(-1) = 1. RHL: h(k-1+1-1) = h(-1+k) as k\to0^+... both give g(-1) = 1. Limit = 1.
Correct Answer: 1