MATEMATİK I

Since the constant function takes the same value for all x ∈ R, we have lim_x→5 7 = 7.

The function is a rational function. Because the limits of the numerator and the denominator are equal to 0, we need to factorise the ratio which follows as

[(x² -2x - 35) / (x - 7)] = [(x - 7) (x + 5)] / x-7 = x + 5. 

Following the limit rules, we find lim_x→7 x + 5 = 7 + 5 = 12.

From the quotient rule for limits together with  lim_x→1 (2x - 8) ≠ 0, we obtain

(1² + 5 . 1 - 12) / (2 . 1 - 8) = -6 / -6 = 1.

From the product rule for limits, we have

lim_x→2 (x + 3) (2x - 5) = lim_x→2 (x + 3) . lim_x→2 (2x - 5)

                              = (2 + 3) . (4 - 5)

                              = -5

Because x approaches -3 from the right side x is more than -3 and for x > -3, f is given by the rule f (x)=–x+12. Hence, the result is as follows. 

lim_x→-3⁺   f(x) = -(-3) + 12 = 3 + 12 = 15

Since the denominator of the expression at x = 2 is 0, we must look at the one-sided limits in order to find the limit at x = 2 if it exists. Then we have,

lim_x→2⁺  (  x  / |x-2| ) =  lim_x→2⁺  (  2  / 0⁺ ) = +∞

lim_x→2^-  (  x  / |x-2| ) =  lim_x→2^-  (  2  / 0⁺ ) = +∞

Hence, since the one-sided limits show the same behavior at x = 2, the limit exist at x =2, and it is +∞.

lim_x→5^- ( x-7 / x - 5) = -2 / 0^- = +∞.

lim_x→5⁺ ( x-7 / x - 5) = -2 / 0⁺ = -∞.

The limit of the function at x = 5 does not exist due to the different behaviour of one-sided limits.

The function is a rational function. Since the degree of the numerator is less than the degree of the denominator, the value of the limit is 0.

f (x) = x² -1 is a polynomial function. Hence, it is a continuous function, and its set of continuity is R.

Since a rational function f is continuous on the domain of its definition, f is continuous on its domain. This function is continuous for all real numbers except 3 which makes its denominator 0. Hence, the set of continuity of this function is R \ {3}.

From the quotient rule for limits together with  we obtain

 The answer is B.

From the quotient rule for limits together with  we obtain

 The answer is D.

From the product rule for limits, we have 

. The answer is B.

Because x approaches 1 from the left side x is less than 1 and for x<1, f is given by the rule f(x)=-2x+5. So, the result follows as

. The answer is E.

Becuase x approaches 3 from the right side x is greater than 3 and for x>3, f is given by the rule . So, the result follows as

. The answer is A.

The answer is B.

Since x approaches 0 from the left side, x is negative and for x<0, |3x|=-3x is true. So, we get following result.

. The answer is C.

The function is a rational function. Since the degree o the numerator is greater than the degreeof the denominator, the following result is true:

. The answer is A.

The function is a rational function. Since the degree of the numerator is less than the degree of the denominator, the following result is true:

. The answer is E.

Since a rational function f is continuous on the domain of its definition, f is continuous on its domain. the  is continuous for all real numbers x except -1 and 1. Thus, f is continuous on             R\{-1, 1}. The answer is B.