# Thread: Finding functions by a given Limits!

1. ## Finding functions by a given Limits!

Hello I gotta find functions by a given limits ;

if there is no such functions just tell me by sayin no such function

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f(x):={x+b(x!=0),17(x=0)}

What values of b give values for the limit of f(x) (x==>0) = L

Need b , and L ( Limit)

b doesnt give L for any value of it
b gives L for any value of it
b give L for only b>0
b give L for only b<0
b give L for only b=0
b give L for only b=17
b give L for only b=-17
b=?,L=? (give L= ! when there is no b )

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Need a function that
limit (x=>zero) F(x) = infinity
limit x(=>zero) X*F(x) = zero
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Need a function that
limit (x=>infinity) F(x) = 0
limit x(=> 0 - ) F(x) = 3
limit x(=> 0 + ) F(x) =1
This function can be written in two ways :
for x<0
or x>= 0
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2. I recommend you go back and re-read the problem. What you wrote here, makes no sense. $\displaystyle \lim_{x\to 0} x+ b= b$ so you make that limit equal to "L" by taking b= L. But then b= L can be any number. Did the problem say something about "continuous"?

$\displaystyle \lim_{x\to 0} F(x)= \infty$, $\displaystyle \lim_{x\to 0} xF(x)= 0$. So F must go to infinity as x goes to 0 but not as fast as x itself. How about 1 over x to any power less than 1? Say, $\displaystyle 1/\sqrt{x}= x^{-1/2}$?

$\displaystyle \lim_{x\to \infty} F(x)= 0$, $\displaystyle \lim_{x\to 0^-}F(x)= 3$, $\displaystyle \lim_{x\to 0^+}F(x)= 1$

As you say, do this as two separate problems. First find a function that goes to 0 as x goes to infinity and goes to 1 as x goes to 0. There are many such functions. A fraction with denominator of higher degree than numerator but with constant terms in numerator and denominator the same with work. Then find a function that goes to 3 as x goes to 0 (for x< 0 but if you find one that works for x> 0, just replace x with -x).

3. Originally Posted by HallsofIvy
I recommend you go back and re-read the problem. What you wrote here, makes no sense. $\displaystyle \lim_{x\to 0} x+ b= b$ so you make that limit equal to "L" by taking b= L. But then b= L can be any number. Did the problem say something about "continuous"?

$\displaystyle \lim_{x\to 0} F(x)= \infty$, $\displaystyle \lim_{x\to 0} xF(x)= 0$. So F must go to infinity as x goes to 0 but not as fast as x itself. How about 1 over x to any power less than 1? Say, $\displaystyle 1/\sqrt{x}= x^{-1/2}$?

$\displaystyle \lim_{x\to \infty} F(x)= 0$, $\displaystyle \lim_{x\to 0^-}F(x)= 3$, $\displaystyle \lim_{x\to 0^+}F(x)= 1$

As you say, do this as two separate problems. First find a function that goes to 0 as x goes to infinity and goes to 1 as x goes to 0. There are many such functions. A fraction with denominator of higher degree than numerator but with constant terms in numerator and denominator the same with work. Then find a function that goes to 3 as x goes to 0 (for x< 0 but if you find one that works for x> 0, just replace x with -x).
Hey there,first thank for the help.I couldnt understand what u meant in 1 + 3 but as for 2 its true (1/x^1/2) for second condition ( when multiplaied with x) but not for the first!