How do you find #lim (3x^2+4)/(x^2-10x+25)# as #x->5#?

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Answer 1 · 2017-02-27T00:18:03

See below.

First, plug in #x = 5# to see if it is a problem:

....

#= (3(5)^2+4)/((5)^2-10(5)+25) = "ndef"#, because the denominator is zero :(

We can then see that:

#lim_(x to 5) (3x^2+4)/(x^2-10x+25)#

# = lim_(x to 5) (3x^2+4)/((x-5)^2)#

Clearly this hits #+oo# as #x to 5# !! (NB: It's not a 2-sided limit, which is often an issue.)

As another approach, in these kinda problems, it can be useful to re-state the Origin, so we say that #z = x - 5#, or #x = z+ 5#.

The problem becomes this:

#lim_(z to 0) (3(z+5)^2+4)/((z+5)^2-10(z+5)+25)#

#= lim_(z to 0) (3z^2+30 z + 79)/(z^2)#

#= lim_(z to 0) 3+30 /z + 79/z^2#

Same conclusion :)