Question

Is this limit undefined

limx->0 (5^(x)-1)/x

Explain why or why not?

Answer 1

\[\lim_{x \rightarrow 0}\frac{5^x-1}{x}\]

Answer 2

solve using first principle

Answer 3

First principle?

Answer 4

i solved it but my lecturer says its indeterminate and idk why

yes first principle malevolence

Answer 5

Its indeterminate because its zero/zero. Notice if you plug in zero you get (5^(0)-1)/0 Have you learned L'hospital's rule?

Answer 6

no i never learned L'hospitals rule

i am suppose to solve it using the first principle table

you know like seeing what happens as the limit x approaches a from the right or left...

Answer 7

I know the answer but I don't really know how to do it approaching from left and right :/ I'm sorry. The answer is ln(5) if you are wondering. But I did that with l'hospital's.

Answer 8

http://tutorial.math.lamar.edu/Classes/CalcI/TheLimit.aspx

Answer 9

Oh. Numerically? O.O

Answer 10

when you make a table and for the left you would say -0.9,-0.09,-0.009,-0.0009,-0.00009

and from the right you would say 0.1,0.01,0.001,0.0001,0.00001
plugging in the values

Answer 11

umm i guess...

Answer 12

Yeah, when you plug those in it should approach ln(5) x.x

Answer 13

I don't know how you would know what ln(5) is without some kind of heading because its a non-terminating decimal xP

Answer 14

i want to know if the limit exist like if from the right and left side are they going the same place? i ended up with the same value 1.609 but they approached strangely..

Answer 15

Strangely?

Answer 16

yeah idk...like from the left i got 1.498304,1.597837,1.60827,1.609321,1.609421

Answer 17

and my lecturer says it does not exist, is it because from the left it did not approach the value smooothly on the right it did i got 1.62245,1.61073,1.609567,1.609451,1.609444

Answer 18

anyways malevolence dont stress about it she probably knew we were suppose to solve it using log5 or something

THANKS SO MUCH!

Answer 19

No problem, sorry I can't do more :/

Answer 20

thats okay