Let us recall the Formal Defn. :
# lim_(x to a) f(x)=l," if and only if, "AA epsilon > 0, EE" a "delta > 0#
such that, # AA x, 0 lt |x-a| lt delta rArr |f(x)-l| lt epsilon.#
In our Problem,
#a=6, f(x)=x/4+3, and, l=9/2.#
So, we have to show that,
#AA epsilon gt 0, EE" a "delta gt 0,#
#"s.t., "AA x, 0 lt |x-6| lt delta rArr |(x/4+3)-9/2| lt epsilon.#
Let #epsilon gt 0# be given. Then,
# |f(x)-l| lt epsilon iff |(x/4+3)-9/2|=|x/4-3/2|=|(x-6)/4| lt epsilon,#
# iff 1/4|x-6| lt epsilon iff |x-6| lt 4epsilon................(star).#
Define the #delta" by, "0 lt delta le 4epsilon.#
Therefore, from #(star),# we see that,
given any #epsilon gt o,# we have a #delta, 0 lt delta le 4epsilon,# such that,
# AA x, 0 lt |x-6| lt delta (le 4epsilon) rArr |(x/4+3)-9/2|ltepsilon.#
Therefore, #lim_(x to 6) (x/4+3)=9/2.#
Hence the Proof.
Enjoy Maths.!
