Let [x] denote the greatest integer function & f(x) be defined in a neighbourhood of 2 by
In this we have to find the values of A & f(2) in order that f(x) may be continuous at x = 2.
I thought to use L hopital to find LHL ,but could not use it because of greatest integer function.
For $1<x<2$ you have $[x+1]=2$, so the limit from the left is $$ \lim_{x\to2^-}\frac{(\exp((x+2)\ln4))^{1/2}-16}{4^x-16}= \lim_{x\to2^-}\frac{(\exp((x+2)\ln4))^{1/2}-16}{x-2}\frac{x-2}{4^x-16} $$ Now, $$ \lim_{x\to2^-}\frac{(\exp((x+2)\ln4))^{1/2}-16}{x-2} $$ is the derivative at $2$ of the function $$ g(x)=(\exp((x+2)\ln4))^{1/2} $$ and $$ g'(x)=\frac{\exp((x+2)\ln4)\ln4}{2(\exp((x+2)\ln4))^{1/2}} $$ so $$ g'(2)=\frac{4^4\ln4}{2(4^4)^{1/2}}=8\ln4 $$ Also $$ \lim_{x\to2^-}\frac{4^x-16}{x-2} $$ is the derivative at $2$ of the function $h(x)=4^x$ and $h'(x)=4^x\ln4$, so your limit is $$ \frac{8\ln4}{16\ln4}=\frac{1}{2} $$
For the limit from the right, do the substitution $t=x-2$, so you have $$ \lim_{t\to0^+}A\frac{1-\cos t}{t\tan t}= \lim_{t\to0^+}A\frac{1-\cos t}{t^2}\frac{t}{\tan t}=\dotsb $$
