Under which conditions would the limit as x approaches 7 not exist? Check all that apply. °the graph decreases without bound as x approaches 7 from the right.

°the graph has a vertical asymptote at x=7

°there is a hole in the graph at x=7

°f(7) does not exist

°lim f(x) ≠ lim f(x)
x->7- x->7+​

(I've done this assigment so here are all the answers.)

Answer :

Q1. f'(x)=-20x+4. When x=11, this is f'(11)=-20×11+4=-220+4=-216.

Q2. Lim(x➝4) of x²+3x-1. The expression can be evaluated by direct substitution without constraints: 16+12-1=27.

Q3. The derivative is 7 for all values of x because the function is linear with a constant slope.

Q4. The function is piecewise. We need the conditions that apply when x is close to zero. We need to look at x approaching from the negative side and the positive side. If the limits are not the same the limit does not exist. So, when x<0 we use 7-x². When x is just less than zero, this left-hand limit is 7. When x>0 we use 10x+7. When x is just bigger than zero, this right-hand limit is 7. We are not interested in x=0, we are only interested in the left- and right-hand limits, which happen to be the same. So the limit is 7.

Q5. f'(x)=-3/x². f'(1)=-3.

Q6.  In all limit questions, you can always ignore what happens when x=limit, because it’s the approaches that matter. So x➝2⁻ means look at the graph on the left of the limit. The horizontal line at y=4 means the left limit is 4. x➝2⁺ means look at the graph on the right of the limit. This is where y=-3, so the right limit is -3. The answer is 4 and -3 for left and right limits respectively.

Q7. First find left and right limits: x➝-1⁻: use x<-1: f(x)=x+1, so left limit is 0; x➝-1⁺: use x>-1: f(x)=1-x: so right limit is 1-(-1)=2. The left and right limits are different, so the limit does not exist.

Q8. 1/(x-7). When x=7, this quantity is not defined and this is a vertical asymptote. When x➝7⁻ we can use a value of x slightly less than 7 and see what happens. First try x=6.99, then 1/(x-7)=-100. Now try x=6.999, then we get -1000. So it’s getting larger in magnitude, and it’s negative. An infinitely large negative value is -∞. So the limit as x approaches 7 from the left is -∞.

Q9. Ignore the closed point at (3,7). For x➝3⁻, we are on the sloping line, so the limit is 1.5.

Q10. As x approaches zero the numerator approaches 1. The denominator gets smaller so its reciprocal gets larger, and it can be positive or negative depending on how we approach zero. So we can’t define the limit, so it doesn’t exist.

Q11. Derivative is -(-3/x²)=3/x². When x=-4, this is 3/16, because (-4)²=16.

Q12. x²-2➝-2 when x➝0.

Q13. (x²-16)/(x-4)=(x-4)(x+4)/(x-4)➝x+4 because the common factor x-4 cancels. Note that the quantity does not exist at precisely x=4 but in the limit it approaches x+4=8 when x=4. So the answer is 8 for the limit.

Q14. The derivative is constant -13 and is independent of t, so the instantaneous velocity is -13.

Q15. 1/(x-2)² is always positive, so it doesn’t matter whether we look at the left or right limits. The result as x➝2 is ∞, which is an undefinable quantity. x=2 is a vertical asymptote.

1,2, & 5

Step-by-step explanation:

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