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let $g(x)=\frac{\sin ax}{ax},f(x)=\frac{\sin x}{x} :a\in \mathbb{R}$

we know that : $f≠g$

now why :

$$\lim_{ x \to 0}\frac{\sin x}{x}=\lim_{ x \to 0}\frac{\sin ax}{ax}=1$$

and also :

$$\lim_{ x \to 0}\frac{\tan x}{x}=\lim_{ x \to 0}\frac{\tan ax}{ax}=1$$

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    set $$ax=t$$ in the limits – 2017-01-15
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    Limit of $2$ functions can be same even if the functions are different (they need not even be related as in your examples). $\lim_{ x \to 0}\frac{\sin x}{x} = \lim _{{x\to 0}}{\frac {\ln(x+1)}{x}}=1$ – 2017-01-15
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    You could even use continuity to prove that for **any** function $\;f\;$ that doesn't vanishes identically in some neighborhood of $\;x_0\;$ , and such that $\;\lim\limits_{x\to x_0}f(x)=0\;$ , one gets $\;\lim\limits_{x\to x_0}\frac{\sin f(x)}{f(x)}=1\;$ – 2017-01-15

2 Answers 2

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Note that $f(x) \neq g(x)$, but we do have that $g(x)=f(ax)$.

So you are asking why $$\lim_{x \to 0 } f(ax)=\lim_{x \to 0} f(x)$$

However, note that if we set $ax=t$, Then $$\lim_{x \to 0 } f(ax)=\lim_{t \to 0 } f(t)=\lim_{x \to 0} f(x)$$

As $ax \to 0$ as $x \to 0$.

This could similarly be done for $\frac{\tan x }{x}$.

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    why ? As $ax \to 0$ as $x \to 0$. – 2017-01-15
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    @Almot1960 remember limit rules: $\lim_{x\to 0} ax = a \lim_{x\to 0} x = a(0) = 0$ for any constant $a$. – 2017-01-15
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Two functions are different means in a given domain they have different value for at least one particular point.or they have different domain.but the limit may be equal as it determines the nature of the curve at that very particular point. and that limit can be the same As for example, $f(x)= x$ ,$x\in[-1,1]$ And $g(x)=-x$, $x\in[-1,1]$ Are different functions . But, $$\lim_{x\to0}f(x)= \lim_{x\to0}g(x)=0$$ As both functional values tend to $0$ as independent variable $x\to0$.