Question

can u plz solve this

Answer 1

Answer:

5. Since x is approaching from a number less than 2, the modulus function is defined such that |x-2|=2-x, x<2

Since the limit results in indeterminate form, L'hopital rule applies:

$\lim_{x \to \-2^-} \frac{|x-2|}{x-2}$ = d/dx(2-x) / d/dx(x-2) = -1/1= -1 QED.

6. I just realise that not all indeterminate form can be solved using L'hopital rule. An alternative to solving it ( limit limx→±∞ P(x)/Q(x), where P(x) is a polynomial of degree n and  Q(x) is a polynomial of degree m) is through dividing both numerator and denominator by the highest denominator degree term (i.e. x^2). However, keep in mind that it is in the square root, so technically you are dividing the denominator by -x (hence dividing by x^2 in the square root). Another thing to keep in mind is we are dividing -x, not x, because when we factor out x^2 from x^2-1, we get another factor--- $\sqrt{x^2}$, which is a modulus function. Since x is always negative, the function yields -x. Therefore, dividing by -x on the top and bottom yields the function $\frac{2+\frac{3}{x} }{-\sqrt{1-\frac{1}{x^2} } }$

Evaluating the limit should give you -2

7. f'x=4x+3

Therefore, f'(0)+3f'(-1)=4(0)+3+3(4(-1)+3)=3+3(-1)=0 QED.

Step-by-step explanation: