Help with Calc problem, limits

  • Thread starter h_k331
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In summary: Anyway, by synthetic division:\mathop\lim\limits_{x\to 1}\frac{-(x^3-2x+1)}{x-1}=\mathop\lim\limits_{x\to 1}\frac{-(x^2+x-1)(x-1)}{x-1}=\mathop\lim\limits_{x\to 1}-(x^2+x-1)=-1
  • #1
h_k331
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This shouldn't be too hard but I'm stuck, brain is getting fried.

lim [(1+2x-x^3)-(1+2(1)-(1)^3)]/(x-1)
x->1

have gotten this far and I'm stuck:

lim (-x^3+2x-1)/(x-1)
x->1

Any help is appreciated, thx.

hk

ETA: fixed typo
 
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  • #2
-x^3 + 2x - 1 = x - x^3 + x - 1 = x(1 - x^2) + x - 1

continue
 
  • #3
working...

hk
 
  • #4
I really really want to just cancel x-1 from top and bottom! But knowing I can't, do I need to multiply top and bottom by something to change the denominator?

hk
 
  • #5
h_k331 said:
I really really want to just cancel x-1 from top and bottom! But knowing I can't, do I need to multiply top and bottom by something to change the denominator?

hk
The numerator has what you need. I just gave you one step in the right direction. Look for another step.
 
  • #6
Well, I gave up and solved it with the [f(a+h)-f(a)]/h formula instead of
[f(x)-f(a)]/(x-a). I was able to get the correct solution of -1 (according to the back of the book), but I'm still curious where I was going wrong with my algebra on the first try. Thanks for your help Dan, I appreciate it.

hk
 
  • #7
[tex] -x^3 + 2x - 1 = x - x^3 + x - 1 = x(1 - x^2) + x - 1=x(1+x)(1-x)-(1-x)=(1-x)[x(1+x)-1] [/tex]

And you can see that the (1-x) has been factored.Simplify with the denominator and then take the limit.

Daniel.
 
  • #8
dextercioby said:
[tex] -x^3 + 2x - 1 = x - x^3 + x - 1 = x(1 - x^2) + x - 1=x(1+x)(1-x)-(1-x)=(1-x)[x(1+x)-1] [/tex]

And you can see that the (1-x) has been factored.Simplify with the denominator and then take the limit.

Daniel.

Doh! Difference of two squares, I tend to miss a lot of those when one of the squares is 1. Thanks Daniel.

hk
 
  • #9
h_k331 said:
I really really want to just cancel x-1 from top and bottom! But knowing I can't, do I need to multiply top and bottom by something to change the denominator?

hk

I just wish to clear up something for the record:

When taking the limit, we are considering values of x close to 1 but NOT equal to 1. Thus it is possible to divide by (x-1). Not trying to upstage anyone here. Anyway, by synthetic division:

[tex]\mathop\lim\limits_{x\to 1}\frac{-(x^3-2x+1)}{x-1}=\mathop\lim\limits_{x\to 1}\frac{-(x^2+x-1)(x-1)}{x-1}=\mathop\lim\limits_{x\to 1}-(x^2+x-1)=-1[/tex]

I know, I'm such a Johnny-come-lately.
 

What is a limit in calculus?

A limit in calculus is a fundamental concept that describes the behavior of a function as its input (x-value) approaches a certain value. It represents the value that the function approaches, but may never quite reach, as the input gets closer and closer to the specified value.

How do you solve limit problems?

There are several methods for solving limit problems, including algebraic techniques, graphing, and using theorems such as the squeeze theorem or the intermediate value theorem. The most common approach is to use algebraic manipulation and substitution to evaluate the limit, but the appropriate method will depend on the specific problem at hand.

What is the significance of limits in calculus?

Limits are essential in calculus because they allow us to study the behavior of functions and understand how they change over time or in response to different inputs. They are also crucial in determining the continuity and differentiability of functions, which are fundamental concepts in calculus.

What is the difference between left and right limits?

Left and right limits are types of one-sided limits, which describe the behavior of a function as the input approaches a given value from either the left or right side. A left limit is the value that the function approaches from the left side of the specified value, while a right limit is the value that the function approaches from the right side.

How do you find the limit of a piecewise function?

To find the limit of a piecewise function, you need to evaluate the left and right limits at the specified value and then determine if they are equal. If the left and right limits are equal, then that is the limit of the function at that point. If they are not equal, then the limit does not exist at that point.

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