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Edward decided to challenge himself to a limit problem. He wanted to

construct a function so that limπ‘₯β†’4
β„Ž(π‘₯) = 2 with the following restrictions.
π‘‡β„Žπ‘’ π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘› 𝑖𝑠 π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘›π‘Žπ‘™, 𝑖𝑒 β„Ž(π‘₯) =𝑓(π‘₯) / 𝑔(π‘₯)
π‘‡β„Žπ‘’ π‘™π‘–π‘šπ‘–π‘‘ π‘šπ‘’π‘ π‘‘ π‘π‘œπ‘›π‘‘π‘Žπ‘–π‘› π‘Žπ‘› π‘–π‘›π‘‘π‘’π‘‘π‘’π‘Ÿπ‘šπ‘–π‘›π‘Žπ‘‘π‘’ π‘“π‘œπ‘Ÿπ‘š, ie β„Ž(0) =𝑓(0) / 𝑔(0) = 0 / 0
π‘‡β„Žπ‘’ π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘› 𝑓(π‘₯) / 𝑔(π‘₯) π‘šπ‘’π‘ π‘‘ π‘π‘œπ‘›π‘‘π‘Žπ‘–π‘› π‘Ž π‘Ÿπ‘Žπ‘‘π‘–π‘π‘Žπ‘™ 𝑖𝑛 π‘‘β„Žπ‘’ π‘›π‘’π‘šπ‘’π‘Ÿπ‘Žπ‘‘π‘œπ‘Ÿ
π‘Žπ‘›π‘‘ π‘Ž π‘‘π‘Ÿπ‘–π‘›π‘œπ‘šπ‘–π‘Žπ‘™ 𝑖𝑛 π‘‘β„Žπ‘’ π‘‘π‘’π‘›π‘œπ‘šπ‘–π‘›π‘Žπ‘‘π‘œπ‘Ÿ so that the technique of rationalization can be used
to solve the limit.
Determine such a function.

since h(4) β†’ 0/0, you will need x-4 in the top and bottom

So, I'd start with
h(x) = (x^2-16)/((x-4)(x-3))
since the x-3 factor becomes just 1, having no real impact.
as written, h(x) = (x+4)/(x-3) so h(4) = 8
Now, you want a radical up top, so if we make it come out to something squared, it will be simple.
h(x) = (√(3x-8)*(x^2-16)) / ((x-4)(x-3))
now h(4) β†’ 16, so put 8 in the bottom
h(x) = (√(3x-8)*(x^2-16)) / (8(x-4)(x-3))
Now h(4) β†’ 2