Calculus Help with Limits (check my work please)

Now, if you can figure out the limits of the numerator and denominator, you should be able to figure out the limit of the fraction.
  • #1
NinjaLink
3
0
[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.)
 
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  • #2
NinjaLink said:
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"?
 
  • #3
Will problem 4 be x = negative infinity and x = negative infinity?
 
  • #4
hello?
 
  • #5
NinjaLink said:
Will problem 4 be x = negative infinity and x = negative infinity?
Close, but no cigar.
NinjaLink said:
hello?
Hi there :-p
 
  • #6
NinjaLink said:
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!
 

Related to Calculus Help with Limits (check my work please)

1. What is a limit in calculus?

A limit in calculus is the value that a function approaches as its input (x-value) gets closer and closer to a certain value. It represents the behavior of a function near a specific point, rather than at that point.

2. How do I find the limit of a function?

To find the limit of a function, you can either use algebraic methods or graphing methods. Algebraically, you can plug in values closer and closer to the given point and see what value the function approaches. Graphically, you can use a graphing calculator or sketch the graph of the function to estimate the limit.

3. What are the common types of limits in calculus?

The common types of limits in calculus include: one-sided limits, where the function approaches a certain value from either the left or the right side; infinite limits, where the function approaches positive or negative infinity; and indeterminate limits, where the function approaches a value that cannot be determined without further calculations.

4. How do I know if my limit calculation is correct?

To check if your limit calculation is correct, you can use the definition of a limit and see if your result matches the definition. Additionally, you can use a graphing calculator to graph the function and see if your calculated limit matches the y-value of the point on the graph. You can also ask a teacher or tutor to review your work.

5. How can I improve my understanding of limits in calculus?

To improve your understanding of limits in calculus, it is important to practice solving various types of limit problems. You can also watch online tutorials or attend review sessions to reinforce key concepts. It may also be helpful to create study aids, such as flashcards or summaries, to review the main ideas and formulas associated with limits.

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