Question

( sqrt{x+6}-sin (x-3)-3 )
( x rightarrow 3=(x-3) cos (x-3) )
Umit drectly we Applyng
bra get
So, go for L-Hopital Rule,
ie, Differentiale both numerator &
then apply bmity denonuinator,
( left.Rightarrow frac{1}{i}right)^{n} frac{d}{d x}(sqrt{n+6}-sin (x-3)-3] )
( =frac{1}{2 sqrt{x+6}}-cos (x-3) )
ii) ( frac{d}{d x}[(x-3) cos (x-3)) )
( =frac{(x-3)[-sin (x-1)]-cos (x-3)}{1} )
( therefore frac{u}{x rightarrow 3} frac{sqrt{x+6}-sin (x-3)-3}{(x-3) cos (x-3)}=frac{4}{x rightarrow 3} frac{1}{2 sqrt{x+6}}-cos (x-3) )
( frac{2 frac{1}{2 sqrt{9}}-frac{cos 0^{-}}{2.3}-1}{0-cos 0}=frac{1}{-1}=frac{1}{frac{5}{6}}=frac{1}{frac{1}{1}} )

# Tim Vx+6 - sin(x-3)-3 273 (x-3) cos(x-3) is equal to : D

Solution