JEE Main 2023 — Mathematics Calculus

Given,

Function f(x)={\begin{matrix}\frac{x}{|x|}, & x\neq 0 \\ 1, & x=0\end{matrix},g(x)={\begin{matrix}\frac{\mathrm{sin}(x+1)}{(x+1)}, & x\neq -1 \\ 1, & x=-1\end{matrix} and $h(x)=2[x]-f(x)$h(x)=2[x]−f(x)

To find $\underset{x\rightarrow 1}{\mathrm{lim}}g(h(x-1))$x→1lim​g(h(x−1))

Taking $L.H.L$L.H.L, we take $x=1-h$x=1−h

$$\underset{h\rightarrow {0}^{-}}{\mathrm{lim}}g(h(1-h-1))$$h→0−lim​g(h(1−h−1))

$$\Rightarrow \underset{h\rightarrow {0}^{-}}{\mathrm{lim}}g(2[-h]-\frac{-h}{|-h|})$$⇒h→0−lim​g(2[−h]−∣−h∣−h​)

$$\Rightarrow \underset{h\rightarrow {0}^{-}}{\mathrm{lim}}g(2(-1)+1)$$⇒h→0−lim​g(2(−1)+1)

$$\Rightarrow \underset{h\rightarrow {0}^{-}}{\mathrm{lim}}g(-1)=1$$⇒h→0−lim​g(−1)=1

Now taking $R.H.L$R.H.L, we take $x=1+h$x=1+h

$$\underset{h\rightarrow {0}^{+}}{\mathrm{lim}}g(h(1+h-1))$$h→0+lim​g(h(1+h−1))

$$=\underset{h\rightarrow {0}^{+}}{\mathrm{lim}}g(2[h]-f(h))$$=h→0+lim​g(2[h]−f(h))

$$=\underset{h\rightarrow {0}^{+}}{\mathrm{lim}}g(0-1)$$=h→0+lim​g(0−1)

$$\Rightarrow \underset{h\rightarrow {0}^{+}}{\mathrm{lim}}1=1$$⇒h→0+lim​1=1