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Find the limit if it exist
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\[\lim_{x \rightarrow \infty}\frac{ 5+3x-x^2 }{ 4x^2-25 }\]
consider -1x^2/4x^2 =-1/4
There is a break of the continuous graph in both x = 2.5 and x = -2.5 This can be solved looking at the denominator 4x^2-25 (2x-5)(2x+5) Now solve for 0 2x-5 = 0 2.5 = x 2x+5 = 0 -2.5 = x
Does that make sense?
nope
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denominator = 0 is not a limit though, that is when the equation is undefined
Correct. But it does show where the limit of the graph is. When the graph approaches -2.5 and 2.5, it will reach to -infinity and +infinity. From the information you will be able to see the limits of it.
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