evaluate the limit $\lim_{x\to 2+}\frac{[x]\sin (x-2)}{(x-2)^2}$

Question

evaluate this limit $$\lim_{x\to 2+}\frac{[x]\sin (x-2)}{(x-2)^2}$$, where $[x]$ represents fractional part of x.

$[x]=x-${$x$}, where {.} is intergral part of x.

my doubt is when $${x\to 2+}$$ then,

$$\frac{\sin (x-2)}{(x-2)}{\to 0}$$ so, $${[x]\to 0.0000...01}$$

then, in denominator the other $(x-2)$ will tend to zero or $${(x-2)\to 0.0000...01}$$

Answer 1

Do the substitution $t=x-2$, which makes things much easier, by noticing that the fractional part of $t+2$ is equal to the fractional part of $t$ and, when $0<t<1$, the fractional part of $t$ equals $t$. So your limit is $$ \lim_{t\to0^+}\frac{t\sin t}{t^2}=1 $$

Answer 2

Certain facts you should know to answer question:

$$\lim_{x\rightarrow 0}\frac{\sin(x)}{x}=1$$ $$\lim_{x\rightarrow2^{+}}[x]=0$$ Since $\lim_{x\rightarrow2^{+}}x-2=0$

So,$$\lim_{x\rightarrow2^{+}}\frac{\sin(x-2)}{(x-2)}*\frac{[x]}{(x-2)}$$ $$=\lim_{x\rightarrow2^{+}}1*\frac{[x]}{(x-2)}$$ For $x\rightarrow2$, {x}=2. So it simplifies to $$\lim_{x\rightarrow2^{+}}\frac{(x-2)}{x-2}$$ $$=1$$