For which of these functions f(x) does limit as x approaches negative infinity f(x)=2

A. (x-2)/(3x-5)
B. 2x/sqrt(x-2)
C. (2x^2-6x+1)/(1+x^2)
D. (2x-1)/(x^2+1)
E. None of these

To determine which of these functions has a limit of 2 as x approaches negative infinity, we can evaluate the limit of each function separately.

Let's analyze each option:

A. To determine the limit as x approaches negative infinity, we divide the leading term of the numerator (x) by the leading term of the denominator (3x). The result is 1/3. Therefore, option A does not have a limit of 2 as x approaches negative infinity.

B. We divide the leading term of the numerator (2x) by the leading term of the denominator (√(x-2)). Here, we have an issue. The denominator contains a square root of (x-2), which becomes undefined for negative values of x. Since the function is undefined for negative values of x, option B does not have a limit as x approaches negative infinity.

C. This function has the highest exponent in the numerator and denominator as x approaches negative infinity. By dividing the leading term of the numerator (2x^2) by the leading term of the denominator (x^2), we get 2. Hence, option C has a limit of 2 as x approaches negative infinity.

D. We divide the leading term of the numerator (2x) by the leading term of the denominator (x^2). This yields 0. Therefore, option D does not have a limit of 2 as x approaches negative infinity.

E. We have already identified that option C, (2x^2-6x+1)/(1+x^2), has a limit of 2 as x approaches negative infinity. Thus, option E is incorrect.

In conclusion, the function that has a limit of 2 as x approaches negative infinity is option C, (2x^2-6x+1)/(1+x^2).

To determine which of these functions have a limit of 2 as x approaches negative infinity, we can evaluate each option.

A. To find the limit of (x-2)/(3x-5) as x approaches negative infinity, we divide both the numerator and denominator by x:

lim(x->-∞) [(x-2)/(3x-5)]
= lim(x->-∞) [1 - (2/x)] / (3 - (5/x))

As x approaches negative infinity, we have 2/x approaching 0, so the limit becomes:

= [1 - 0] / (3 - 0)
= 1/3

Since the limit is not equal to 2, option A is not correct.

B. For the function 2x/sqrt(x-2), as x approaches negative infinity, both the numerator and denominator go to negative infinity. We can simplify the function by dividing both the numerator and denominator by x:

lim(x->-∞) [2x/sqrt(x-2)]
= lim(x->-∞) [2 / (√(1 - (2/x))]

As x approaches negative infinity, we get 2/0, which is undefined. Therefore, option B is not correct.

C. For the function (2x^2-6x+1)/(1+x^2), as x approaches negative infinity, both the numerator and denominator go to positive infinity. Dividing both the numerator and denominator by x^2 yields:

lim(x->-∞) [(2-∙6/x+1/x^2) / (1/x^2 + 1)]

As x approaches negative infinity, we have 6/x and 1/x^2 approaching 0, so the limit becomes:

= (2 - 0 + 0) / (0 + 1)
= 2/1
= 2

Since the limit is equal to 2, option C is correct.

D. For the function (2x-1)/(x^2+1), as x approaches negative infinity, both the numerator and denominator approach negative infinity. We can divide both the numerator and denominator by x^2:

lim(x->-∞) [(2/x - 1/x^2) / (1 + 1/x^2)]

As x approaches negative infinity, we have 2/x and 1/x^2 approaching 0, so the limit becomes:

= (0 - 0) / (1 + 0)
= 0/1
= 0

Since the limit is not equal to 2, option D is not correct.

Therefore, the correct answer is C. (2x^2-6x+1)/(1+x^2)