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$$1-)\lim _{x \to 0}\frac{\sin x^2 }{x}$$

$$2-)\lim _{x \to 0}\frac{\sin x^2 }{x}\times \frac{x}{x}$$

$$3-)\lim _{x \to 0}\frac{\sin x^2 }{x^2}\times x$$

$$4-)\lim _{x \to 0}\frac{\sin x^2 }{x^2}\times\lim _{x \to 0} x=1\times0=0$$

now : in Step (2) why We can $\frac{\sin x^2 }{x}$in $\times \frac{x}{x}$?

We know that if be $x=0$ then $1/0=$Undefined

Generally: prove that :

if $$g(a)=0$$

then :

$$\lim _{x \to a}f(x)=\lim _{x \to a}\frac{f(x)\times g(x)}{g(x)}$$

thank you .

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    In general if $\lim_{x\to a}g(x)$ exists then you can do what you did. – 2017-02-12
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    @kingW3: this has nothing to do with limit of $g(x) $ as $x\to a$. – 2017-02-13

1 Answers 1

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A limit is the value a function approaches as $x$ approaches some point. As long as it is true that

$$f(x)=f(x)\frac{g(x)}{g(x)}$$

for $x$ near the point it is approaching (but not necessarily at the point), then we may have

$$\lim_{x\to a}f(x)=\lim_{x\to a}f(x)\frac{g(x)}{g(x)}$$

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    Thus as long $g(x) \neq 0$ as $x\to a$ the desired equation holds. This is independent of the the fact whether limit of $g(x) $ as $x\to a$ exists or not. +1 – 2017-02-13