- #1

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## 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

- Thread starter goodheavens
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- #1

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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

- #2

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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:

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:

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- #3

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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

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

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

- #6

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i see it now. thank you :)

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