# Limit of a composite function

## Homework Statement

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

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

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

$$\lim_{t \to 0} \frac{sin(t)}{t} =1$$

There is a geometric proof of the above on youtube:

Last edited by a moderator:
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

SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

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

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 $$\displaystyle g(x)=\left\{\begin{array}{cc}\displaystyle {{\sin x}\over{x}},&\mbox{ if } x\neq 0\\ \\ 1, & \mbox{ if } x=0\end{array}\right.$$

is continuous on $$\mathbb{R},$$ the set of all real numbers.

Proving that the limit of sin(t)/t is 1 as t->0 is easy, just expand sin(t) as a series.

i see it now. thank you :)