Question

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

Answer 1

`ln3^9`

Explanation:

`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

`9ln3`

Explanation:

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."`

Power rule:

`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`