What's the limit of this function?

[lillybeans]

Homework Statement

Find limit of x --> 3 of

ln(1+2x)-ln7
-----------
x-3

Homework Equations

The properties of log?

The Attempt at a Solution

= ln ((1+2x)/7)^(1/x-3)

then make h=x-3; x=h+3

limit as h --> 0

= ln ((1 + 2h + 6)/7) ^ (1/x-3)
= ln ((7 + 2h)/7) ^ (1/x-3)
= ln (1 + 2/7h)^(1/x-3)

then I did it the non-algebraic way and plugged in 0.01 and got 0.2587 as the answer, which is coincidentally 2/7 (this number is found in the equation also)! But how do I prove/arrive at that? I do remember that (1+h)^(1/h) = e. But what do I do now?

Help is appreciated and many thanks.

 

[micromass]

Hi lillybeans!

Isn't the easiest thing to apply l'Hopitals rule on

$$\lim_{x\rightarrow 3}{\frac{ln\left(\frac{1+2x}{7}\right)}{x-3}}$$

 

[Hurkyl]

limit as h --> 0

= ln ((1 + 2h + 6)/7) ^ (1/x-3)
= ln ((7 + 2h)/7) ^ (1/x-3)
= ln (1 + 2/7h)^(1/x-3)

It would help if you changed all of your x's to h's. (And a nitpick -- it would help if you used parentheses correctly in the denominator of the fraction of your exponent)

I do remember that (1+h)^(1/h) = e. But what do I do now?

That's almost what you have already, isn't it?

 

[micromass]

Your method could work too, actually. Just put u=(2/7)h and see what the substitution gives you...

 

[lillybeans]

It would help if you changed all of your x's to h's. (And a nitpick -- it would help if you used parentheses correctly in the denominator of the fraction of your exponent)


That's almost what you have already, isn't it?

Ahhh right, I'm sorry, I forgot to change the exponent for some reason, when I did it on paper I DID change 1/x-3 to 1/h.

@ Micromass: Thank you for your help! I'm not too familiar with l'hopital's rule but I will look it up. Also, thank you for the hint! I guess I didn't think of substituting variables TWICE (first h, now u) :)