What is the limit of log10 (x2-5x+6) as x approaches 3?

[shorti2406]

I guess I am having trouble understanding this problem. I don't even really know where to start, or what rules to use.

Find the Limit:

lim log10 (x2-5x+6)
x->3+

 

[geniusprahar_21]

lim log (x2 - 5x +6)
x->3+

Put x = 3+h, where h>0 and limit of h as zero solving, we have

lim log (h2+h) = lim logh + lim log(h+1) = lim log h = (-infinity)
h->0 h->0 h>0 h->0

 

[M98Ranger]

To find the limit of a function as x approaches a certain value, we can use the direct substitution method or algebraic manipulation. In this case, we can use algebraic manipulation to find the limit of log10 (x2-5x+6) as x approaches 3.

First, we can factor the expression inside the logarithm to get (x-3)(x-2). Then, we can rewrite the expression as log10 [(x-3)(x-2)].

Next, we can use the logarithmic property loga (xy) = loga (x) + loga (y) to rewrite the expression as log10 (x-3) + log10 (x-2).

Now, as x approaches 3, both (x-3) and (x-2) approach 0. This means that log10 (x-3) and log10 (x-2) both approach negative infinity. Using the logarithmic property loga (x) = -infinity as x approaches 0, we can rewrite the expression as -infinity + -infinity, which equals -infinity.

Therefore, the limit of log10 (x2-5x+6) as x approaches 3 is -infinity. This means that the value of the function approaches negative infinity as x gets closer and closer to 3.