Calculus Help with Limits (check my work please)

[NinjaLink]

[SOLVED] Calculus Help with Limits (check my work please)

Some problems, I just couldn't do on my own, but I've done some problems with answers provided here. Please check my work and show me how to do the problems I couldn't do. I would appreciate it.


Problem 1:

Lim
x--> 3^- (left side)
x--> 3^+ (right side)
x--> 3

Expression:

|x+3| / (x-3) <== absolute value expression

--------------------------

Problem 2:

lim x--> 3^-(left side)

f(x) , where

5x-7 if x is less than or equal to 3
and
x^2+1 if x is greater than 3

Answer: f(x) = 8 (I'm not sure if it is right. Correct me if I'm wrong)

---------------------------

Problem 3:

lim
x-->3

(x^2-2x-3)/(x^2-6x+9)

Answer: 1 (I'm not sure if it is right. Correct me if I'm wrong)

---------------------------

Problem 4:

lim
x--> 2^- (left side)
x--> 2^+ (right side)

(3x+1)/(x-2)

--------------------------

Problem 5:

lim
x-->3

(2x+6)/(x)

Answer: 4 (I couldn't figure it out whether it was 4 or 8.)

 

[HallsofIvy]

Some problems, I just couldn't do on my own, but I've done some problems with answers provided here. Please check my work and show me how to do the problems I couldn't do. I would appreciate it.


Problem 1:

Lim
x--> 3^- (left side)
x--> 3^+ (right side)
x--> 3

Expression:

|x+3| / (x-3) <== absolute value expression

What happens if x is close to 3? Suppose x= 3.001 or x= 2.999. What is |x+3|/(x-3) in each of those cases? What happens if x is even closer to 3?

--------------------------

Problem 2:

lim x--> 3^-(left side)

f(x) , where

5x-7 if x is less than or equal to 3
and
x^2+1 if x is greater than 3

Answer: f(x) = 8 (I'm not sure if it is right. Correct me if I'm wrong)

If x< 3 which is what "3^-" means, then f(x)= 5x-7. That's linear so continuous so the limit is just f(3)= 5(3)- 7= 8.

---------------------------

Problem 3:

lim
x-->3

(x^2-2x-3)/(x^2-6x+9)

Answer: 1 (I'm not sure if it is right. Correct me if I'm wrong)

That's a rational function so it's continuous wherever the denominator is not 0. Does that happen here? if x= 3, it is (3²- 2(3)-3)/(3²- 6(3)+ 9)= (9- 6- 3)/(9- 18+ 9)= 0/0. We can't just set x= 3 and find the limit. But the fact that both numerator and denominator are 0 at x= 3 tells us that x- 3 is a factor of each:
(x²- 2x- 3)/(x²- 6x+ 9)= (x- 3)(x+ 1)/(x- 3)(x- 3). As long as x is not 3, we can cancel the two (x-3) factors: (x+1)/(x- 3). Now what is the limit of that as x goes to 3?

---------------------------

Problem 4:

lim
x--> 2^- (left side)
x--> 2^+ (right side)

(3x+1)/(x-2)

Again, what happens if x is very close to 2? What if x= 2.0001? What if x= 1.999?
Notice that in both this and problem 1, the denominator is 0 at the "target" point but the numerator is not

--------------------------

Problem 5:

lim
x-->3

(2x+6)/(x)

Answer: 4 (I couldn't figure it out whether it was 4 or 8.)

What is that fraction equal to when x= 3? Why would you even think "8"?

 

[NinjaLink]

Will problem 4 be x = negative infinity and x = negative infinity?

 

[NinjaLink]

hello?

 

[Hootenanny]

Will problem 4 be x = negative infinity and x = negative infinity?

Close, but no cigar.

hello?

Hi there

 

[HallsofIvy]

Will problem 4 be x = negative infinity and x = negative infinity?

No, problem 4 tells you that x will be close to 3. It doesn't ask for values of x!