Question

When x=1,2,3,.... find limit x->infinity

choice
a. 0
b. 1
c. 2
d. 3
f. 4
help me!!!
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Answer 1

Let y = √x, so the limit can be rewritten as

`\displaystyle \lim_(y\to\infty)\left((√(y^6+y^2))/(y^2+3) - (√(y^4+1))/(y+4)\right)`

Now,

`√(y^6+y^2) = √(y^2\left(y^4+1\right)) = y√(y^4+1)`

so we can rewrite the limit further as

`\displaystyle \lim_(y\to\infty)\left((y)/(y^2+3) - \frac1{y+4}\right)√(y^4+1)`

Combine the rational terms:

`\frac y{y^2+3} - \frac1{y+4} = (y(y+4)-(y^2+3))/((y^2+3)(y+4)) = (4y-3)/((y^2+3)(y+4))`

Then in the limit, we get

`\displaystyle \lim_(y\to\infty)((4y-3)√(y^4+1))/((y^2+3)(y+4)) = \lim_(y\to\infty)\frac{(4y^3-3y^2)\sqrt{1+\frac1{y^4}}}{y^3+4y^2+3y+12} \\\\ = \lim_(y\to\infty)\frac{\left(4-\frac3y\right)\sqrt{1+\frac1{y^4}}}{1+\frac4y+\frac3{y^2}+(12)/(y^3)} = \boxed{4}`