Question

How do I get the answer for this

Answer 1

`\bf \lim\limits_(x\to -\infty)~\cfrac{√(9x^2+2)}{2x-9}\implies \cfrac{\lim\limits_(x\to -\infty)~√(9x^2+2)}{\lim\limits_(x\to -\infty)~2x-9}`

now, for any "x" value, the numerator will give you a positive root, say x = -10, then 9(-10)^2 +2 will be 902, square root of that is about 30.

whilst for the denominator, that'd be 2(-100) -9, or -209.

as you can see, the denominator for any "x" value, has larger jumps then the numerator, and because of that the rational itself, will always have a larger and larger and negative denominator.

so it may go from say -1/100 to -1/1000000 and so on as "x" progresses, going ever closer and closer to 0.