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

Answer 1

Aurora Alvarado

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 :)

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Answer 2

Evelyn Agard

To find the limit of (3x^2+4)/(x^2-10x+25) as x approaches 5, we can substitute 5 into the expression and simplify. By substituting x=5, we get (3(5)^2+4)/(5^2-10(5)+25). Simplifying further, we have (75+4)/(25-50+25). Continuing to simplify, we get 79/0. Since the denominator is 0, the limit does not exist.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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

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