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Solving a limit help

  • Thread starter maxitis
  • Start date
  • #1
6
0
I have the limit:
[tex]
\lim_{x\rightarrow +2} {\frac{f(x)-5}{x-2}}=5
[/tex]
And i want to find the:
[tex]
\lim_{n\rightarrow +2} {f(x)}
[/tex]
Can i say that [itex]f(x)-5=5*(x-2)[/itex]
And then find the limit?
Thank you
 
Last edited:

Answers and Replies

  • #2
6,054
390
There is some confusion between n and x. Fix that. Then observe that as x (or n) goes to 2, the denominator goes to zero.
 
  • #3
6
0
And why is that a problem?
Both parts will be equal to 0.
 
  • #4
6,054
390
Because that would prevent you from using the formula you wanted.

But think about it. You have a finite limit of a ratio, whose denominator goes to zero. What can be said about the limit of its numerator in this case?
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,770
911
Yes, the denominator goes to 0. So what must the numerator go to in order that a limit exist?
 
  • #6
HallsofIvy
Science Advisor
Homework Helper
41,770
911
I have the limit:
[tex]
\lim_{x\rightarrow +2} {\frac{f(x)-5}{x-2}}=5
[/tex]
And i want to find the:
[tex]
\lim_{n\rightarrow +2} {f(x)}
[/tex]
Can i say that [itex]f(x)-5=5*(x-2)[/itex]
And then find the limit?
Thank you
No, because there are, in fact, an infinite number of possible functions, f. f(x)= 5(x- 2)+5 is just one of them.

Yes, the denominator goes to 0. So what must the numerator go to in order that a limit exist?
 
  • #7
6
0
The numerator should be zero.
So we have: [itex]f(2)-5=0[/itex] Am i right?

Something about before,
Why you said that there are infinite number of functions f that can satisfy that limit?
One is [itex]f(x)=5(x-1)[/itex], can you give me another example?
Thank you
 
Last edited:

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