$lim_(x->0)(3^(2+x)-9)/x$ 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 $0$

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

$=>9ln3$

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