How to calculate this limit

(x+√(x+3))/(2x-1)
x-> + infinity

Let x + 3 = y
x → +∞ => y → +∞
=> Limit
= lim (y → +∞) (y - 3 + √y) / [2(y - 3) - 1]
= lim (y → +∞) (y - 3 + √y) / (2y - 7)
Dividing the numerator and denominator with y,
limit
= lim (y → +∞) (1 - 3/y + 1/√y) / (2 - 7/y)
= 1/2.

rewritten:

(x + (x+3)^0.5) / (2x-1)

The largest power of x in the numerator is of degree one, (x)
The largest power of x in the denominator is also of degree one, (2x)

Throw out the rest: lim(x->infinity) x/(2x) = 1/2
Regardless of how big x is, x / 2x will equal 0.5.

divide numerator and denominator by x
limit becomes..... ( 1 + root[ 1/x + 3/x^2 ] ) / ( 2 - 1/x )
apply limit x --> +infinity
i.e. 1/x ---> +zero
limit becomes .......... 1/2
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.. lim (x→∞) { [ x + √(x+3) ] / ( 2x - 1 ) }
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... Dividing top & bottom by x or √(x²)
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= lim (x→∞) { [ 1 + √((1/x)+3(1/x²)) ] / [ 2 - (1/x) ] }

= [ 1 + √(0+3(0)) ] / [ 2 - (0) ]

= [ 1 + 0 ] / [ 2 ]

= 1/2 ........................................… Ans.
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The answer is 1/2
You take the largest exponent in the denominator and divide everything by it (x). Anything that is over x is zero. You are then left with 1+0 over 2+0
1/2

quotient rule probably.
OR
since it says to infinity.. plug in like 1,000,000,000 and see what number it is getting close to.
for ex.. 5.0000000000000000000000000012

Because i am only in 8th grade i dont know.