Infinite Limits and Infinity at Limits

Question

1.) Lim

x->4+ ( -(x-2)/(4-x) + (5x)/(2x+3) )

2.) Lim

x-> 3- ( (2x/x-3) - (x+8)/(5x+2) )

3.) Lim

x-> +∞ ( (8x^4-2x^3+x^2+7x-1) / (2x^4+2x^3-3x+11)

4.) Lim

x-> +∞ ( (7x^2+2x-10) / (2x^3-x^2+9x) )

5.) Lim

x-> -∞ ( (-3x^2+12x-1) / (2x+3) )

*note: please provide a complete solution, with step-by-step explanation on how to solve and final answer, so I can clearly understand.. Thank You!*

Answer 1

1)

lim ( -(x-2)/(4-x) + (5x)/(2x+3) ) =

lim ( -(x-2)/(4-x)) + lim((5x)/(2x+3)) because limit of a sum is a sum of limits

lim((-(x-2)/(4-x))

is a limit of a fraction where

the numerator has the value -2 at x=4, and

the denominator has the value 0 at x=4.

Therefore the limit will be +∞ or -∞ depending on the value of the fraction close to 4.

We are to consider only values of x>4, since the limit is x->4+.

For those values of x, the fraction is negative, therefore the limit is -∞.

lim((5x)/(2x+3)) = 20/11,

but its exact value is not important (when added to -∞).

It is important that it is defined and finite, which makes the result -∞.

2.)

It is exactly the same reasoning as above.

lim(2x/x-3) is +∞ or -∞ and

lim (x+8)/(5x+2) is defined and finite, hence becomes irrelevant (when added to ∞).

We are to consider only values of x<3, since the limit is x->3-.

For those values of x, the fraction is negative, therefore the limit is -∞.

3.)

As x->+∞ only the highest order term of any polynomial becomes relevant, therefore

lim ((8x^4-2x^3+x^2+7x-1) / (2x^4+2x^3-3x+11)) =

lim ((8x^4) / (2x^4)) =

lim (4) = 4

4.)

As x->+∞ only the highest order term of any polynomial becomes relevant, therefore

lim ((7x^2+2x-10) / (2x^3-x^2+9x)) =

lim ((7x^2) / (2x^3)) =

lim (7/(2x)) = 0

5.)

As x->-∞ only the highest order term of any polynomial becomes relevant, therefore

lim ((-3x^2+12x-1) / (2x+3)) =

lim ((-3x^2) / (2x)) =

lim ((-3/2)x) = +∞