Limit of a composite function

  • #1

Homework Statement



If lim f(x) as x->0 is = 0 then lim [tex]\frac{sin(f(x))}{f(x)}[/tex] as x->0 = 1?


dont know how to start proving this . thanks for the replies
 

Answers and Replies

  • #2
When you say prove do you mean a rigorous proof as in epsilons and deltas ? Because that would be difficult.

You could make the substitution t = f(x) and applying the following

[tex] \lim_{t \to 0} \frac{sin(t)}{t} =1 [/tex]

There is a geometric proof of the above on youtube:
 
Last edited by a moderator:
  • #3
i've thought of that method also but there's this theorem that we have to use called the theorem on limit of a composite function which states that

if lim g(x) as x->a is = b and if the function f is continuous at b,
lim (f o g) (x) as x->a is = f(b)
or, equivalently,
lim f(g(x)) as x->a is = f(lim g(x)) as x->a
 
  • #4
SammyS
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Homework Statement



If lim f(x) as x->0 is = 0 then [tex]\lim_{x\to 0}\frac{\sin(f(x))}{f(x)}= 1\ ?[/tex]

don't know how to start proving this . thanks for the replies
i've thought of that method also but there's this theorem that we have to use called the theorem on limit of a composite function which states that

if lim g(x) as x->a is = b and if the function f is continuous at b,
lim (f o g) (x) as x->a is = f(b)
or, equivalently,
lim f(g(x)) as x->a is = f(lim g(x)) as x->a

The function [tex]\displaystyle g(x)=\left\{\begin{array}{cc}\displaystyle {{\sin x}\over{x}},&\mbox{ if }
x\neq 0\\ \\ 1, & \mbox{ if } x=0\end{array}\right.[/tex]

is continuous on [tex]\mathbb{R},[/tex] the set of all real numbers.
 
  • #5
351
1
Proving that the limit of sin(t)/t is 1 as t->0 is easy, just expand sin(t) as a series.
 
  • #6
i see it now. thank you :)
 

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