General Doubt

Question

Qus 1

Qus 2

Comments

Q1)applying limit in the form of 0/0 so apply L-hospitals rule for both numerator and denominator independently.

we get $lim_{x->1} -e^{-x}/1$

then apply limit =-1/e

Q2) same applying limit 0/0

after apply L-hospitals rule     $lim_{x->0} (\frac{p}{1+px})/3.e^{3x}$

=p/3