$$ \lim_{x \to 0} \frac{5^x-1}{x} $$ if I use the L'hospital rule, how I do?
is this correct?

$$ \lim_{x \to 0} \frac{ln(5^x-1)=lny}{lnx=lnz} $$

Question

$$ \lim_{x \to 0} \frac{5^x-1}{x} $$  if I use the L'hospital rule, how I do?
  is this correct?

  $$ \lim_{x \to 0} \frac{ln(5^x-1)=lny}{lnx=lnz} $$

anonymous · Answer

the derivative of 5^x is

$$ 5^x \ln 5$$

anonymous · Answer

$$ \lim_{x \to 0} \frac{5^x \ln 5}{1}= \ln 5$$

jpsmarinho · Answer

It's true @Spacelimbus, I'm wrong. Thanks! If, rather than $$5^x $$ was $$x^x$$? I should use logarithm derivation right?

anonymous · Answer

yes but then things get pretty pretty weird, because the derivate of x^x is

$$ x^x (\ln x +1)$$ So you see already, for this expression, if x approaches zero, then this entire term isn't defined.

anonymous · Answer

and then you are stuck with l'Hospitals rule, because the x^x will never decay if you take it's derivative

jpsmarinho · Answer

I understand ^^.  Thanks!