#lim_(x->0)(3^(2+x)-9)/x# Can you please Evaluate this limit ?

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Answer 1 · 2018-02-27T21:02:30

#ln3^9#

#lim_( x-> 0)(3^(2+x)-9)/x=lim_(x->0)(3^2*3^x-3^2)/x=3^2lim_(x->0)(3^x-1)/x=9ln3=ln3^9#
Hint:
#*lim_(h->0)(a^h-1)/h=lna,where, ainR^+ -{1}#

Answer 2 · 2018-02-28T00:23:12

#9ln3#

Since we get #0/0# after plugging in #0# in the place of #x#, we use the L'Hospital's Rule.

The rule states that:

#lim_(x->c)f(x)/g(x)=(f'(c))/(g'(c))# If we get an indeterminate form for #f(c)/g(c)#

#=>lim_(c->0)(3^(2+x)-9)/x=(d/dx(3^(2+x))-d/dx(9))/(d/dx(x))#

Exponent rule:

#d/dx(n^x)=ln(n)*n^x*d/dx(x)# when #n# is a constant.

#"A derivative of a constant is always zero."#

#d/dx(x^n)=nx^(n-1)# when #n# is a constant.

We now have:

#=>(ln3*3^(2+x)*d/dx(2+x)-0)/1#

#=>ln3*3^(2+x)*1#

#=>ln3*3^(2+x)#

#=>3^(2+x)ln3#

Replace #x# with #0#

#=>3^(2+0)ln3#

#=>9ln3#