ok, so if you tried to evaluate it, we would get something/0, which is undefined
not helpful
hmm
the next step would be to try and factor something out
well, I don't see what to factor out
the next try would be to use l'hopital's rule
take derivitive of top and bottom
derivitive of top is [tex] \frac{x}{ \sqrt{x^2+24} } [/tex]
derivitive of bottom is 1
so we get[tex] \frac{\frac{x}{ \sqrt{x^2+24} }}{1} =\frac{x}{ \sqrt{x^2+24} }[/tex]
if we input 1 for x we get
[tex] \frac{1}{ \sqrt{1^2+24} } [/tex]=[tex] \frac{1}{ \sqrt{1+24} } [/tex]=[tex] \frac{1}{ \sqrt{25} } [/tex]=[tex] \frac{1}{5} [/tex]
we know it can't be [tex] \frac{-1}{5} [/tex] because if we evaluated it at x=0 and x=2, we see that it is above 0
the as x->1, [tex] \frac{ \sqrt{x^2+24} -5}{x-1} [/tex]=1/5
