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Given that:

$$\lim_{x\to 0}\frac{f(x)}{x}=1.$$

How do I evaluate $$\lim_{x\to 0}f(x)$$ and $$\lim_{x\to 0}\frac{f(x)}{g(x)} $$ given $\lim_{x\to 0}g(x)$ exists and is non zero.

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    If $\lim_{x\to 0} f(x)/x=1$ and $\lim_{x\to 0} g(x)/x=2$ then $\lim_{x\to 0} f(x)/g(x)=\frac{f(x)/x}{g(x)/x}=1/2$ – 2017-02-16
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    @kingW3: OP states that $\lim_{x\to0} g(x)$ exists and is non-zero, so it would see unlikely that $\lim_{x\to0} \frac{g(x)}{x}$ exists. (Although it is possible it is an error in the OP.) – 2017-02-16
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    OP: If you mean that the limit of $\frac{g(x)}{x}$ is non-zero (and not the limit of $g(x)$ itself), please edit your post to reflect this. – 2017-02-16

4 Answers 4

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It's not a property, you are just multiplying and dividing a function by a number that is not zero(even though it tends to 0), so you are not changing the function.

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    This is not really an accurate, helpful, or even correct answer. – 2017-02-16
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Hint: $\lim_{x\to 0}{f(x)} = \lim_{x\to 0}{x\cdot \frac{f(x)}{x}}$. What do you know about limits of products or quotients?

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    For limits of quotients we can use lim [ f(x) / g(x) ] = lim f(x) / lim g(x) ; if lim g(x) is not equal to zero. – 2017-02-16
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    Yep, and since you'll have already shown that $\lim_{x\to 0}{f(x)}=0$, you'll see that $\lim_{x\to 0}{\frac{f(x)}{g(x)}} = \frac{\lim_{x\to 0}{f(x)}}{\lim_{x\to 0}{g(x)}}=0$ – 2017-02-16
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$$\lim_{x\to 0}f(x)=\lim_{x\to 0}\left[x\cdot\frac{f(x)}{x}\right]=0(1)=0$$ and so, if we write $\lim_{x\to 0}g(x)=L$ then by assumption $L\in\Bbb R$ and $L\neq 0$ so that we get $$\lim_{x\to 0}\frac{f(x)}{g(x)}=\frac{0}{L}=0.$$

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    how do we get this property? $$\lim_{x\to 0}f(x)=\lim_{x\to 0}\left[x\cdot\frac{f(x)}{x}\right]$$ – 2017-02-16
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    The trick is $$f(x)=x\cdot\frac{f(x)}{x}$$ because we are dealing with $x\to 0$ and in here, $x\neq 0$. – 2017-02-16
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    Having $$\lim_{x\to 0}\frac{f(x)}{x}=1.$$ and then given $$\lim_{x\to 0}\frac{g(x)}{x}=2.$$ when we evaluate the limit of $g(x)$ we get $$\lim_{x\to 0}g(x)=\lim_{x\to 0}\left[x\cdot\frac{g(x)}{x}\right]=0(2)=0$$ so why doesn't $$\lim_{x\to 0}\frac{f(x)}{g(x)} $$ also stay 0 if both $f(x)$ and $g(x)$ are 0 – 2017-02-16
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    The limit $\lim_{x\to 0}g(x)$ must be non-zero before you apply the law which says "the limit of the quotient is the quotient of limits" – 2017-02-16
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    In other words $$\lim_{x\to 0}\frac{f(x)}{g(x)}=\frac{\lim_{x\to 0}f(x)}{\lim_{x\to 0}g(x)}$$ does not hold if $\lim_{x\to 0}g(x)=0.$ – 2017-02-16
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    How would we evaluate the limit for $\lim_{x\to 0}\frac{f(x)}{g(x)}$ when $\lim_{x\to 0}\frac{f(x)}{x}=1.$ and $\lim_{x\to 0}\frac{g(x)}{x}=2.$ – 2017-02-16
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    Get the ratio $\frac{f(x)/x}{g(x)/x}$ and you will get 1/2 – 2017-02-16
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    I have work I go for now – 2017-02-16
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    Ok thanks for help. So we can say $\lim_{x\to 0}\frac{f(x)}{g(x)} = \lim_{x\to 0}\frac{f(x)/x}{g(x)/x}$. which would equal $\frac{\lim_{x\to 0}\frac{f(x)/x}}{\lim_{x\to 0}2}$ $\frac{g(x)/x}}$ – 2017-02-16
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    @lakada: you need to learn about rules of algebra of limits and the conditions under which they hold. Once you get to know these basic rules, questions such as these are trivial. – 2017-02-16
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0/0 is undefined. You would have to apply L'Hospital and analyse limx→0 f'(x)/g'(x)

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    How would we apply L'Hospital in this: $\lim_{x\to 0}\frac{f(x)}{g(x)}$ when $\lim_{x\to 0}\frac{f(x)}{x}=1.$ and $\lim_{x\to 0}\frac{g(x)}{x}=2.$ – 2017-02-16