Estimate the limit by substituting smaller and smaller values of h. Round the answer to one decimal place!

lim (5^h)-1
h-0 -------
h

if i sub in h=0.1, 0.01, 0.001,

i get..
h=0.1.... 1.746
h=0.01... 1.622459
h=0.001... 1.6107

so would my answer be 1.6? or 1.7? :/

Question

Estimate the limit by substituting smaller and smaller values of h. Round the answer to one decimal place!

lim      (5^h)-1
h->0  -------
              h

if i sub in h=0.1, 0.01, 0.001,

i get.. 
h=0.1.... 1.746
h=0.01... 1.622459
h=0.001... 1.6107

so would my answer be 1.6? or 1.7? :/

zepdrix · Answer

Hmm

anonymous · Answer

@OpiGeode not sure what you mean?

anonymous · Answer

haha @zepdrix yeah not sure if i did it correctly.. tried to do it like the last one, but not sure if i succeeded :P

mertsj · Answer

Is it getting closer to 1.7 or 1.6?

zepdrix · Answer

Your calculations looks correct so far.
I would try one more h value to convince yourself :)

h=0.00001 or something.

anonymous · Answer

@Mertsj that i'm not too sure of haha

and okay @zepdrix 

so would i get this?

h=0.000001.... 1.6094392

so would it be 1.6?

haha think i'm more sure now lol:P

mertsj · Answer

yes.  1.6

anonymous · Answer

ahh yay! thank you guys!! @Mertsj @zepdrix :)

zepdrix · Answer

Oh they just want 1 decimal? Ok cool looks like you got it ^^

anonymous · Answer

haha yeah:)

thanks for helping!!:)