Understanding the Limit of Natural Log: ln(x^2-16) as x Approaches 4+

[Painguy]

Determine the infinite limit.
lim x->4+ ln(x^2-16)

I know from graphing the equation and doing a table that the limit is -infinity, but my book is saying to do the following.

Let t = x^2-16, Then as x->4+, t->0+, and lim x->4+ ln(x^2-16)=lim t->0+ ln(t) by 3

 

[Gallagher]

I think it is just substituting in t for the function inside the natural log to make it easier to understand. Putting 4 into the equation you get 4²-16=16-16=0 so it is the limit of ln(0). Thinking about ln(0), it is undefined as it represents the number you would need to take e to the power of to get e^x=0. Since this e^x can never reach 0 but only approach it, ln(0) is undefined and you limit is, as you thought, -∞.

Unless I'm totally wrong.

 

[Painguy]

I think it is just substituting in t for the function inside the natural log to make it easier to understand. Putting 4 into the equation you get 4²-16=16-16=0 so it is the limit of ln(0). Thinking about ln(0), it is undefined as it represents the number you would need to take e to the power of to get e^x=0. Since this e^x can never reach 0 but only approach it, ln(0) is undefined and you limit is, as you thought, -∞.

Unless I'm totally wrong.

ah i see. i guess that makes sense. It just seems rather pointless lol.