Let's do some calculus! (14)

Calculus Level 2

If $\displaystyle \lim_{x \to a} {\big(f(x)\cdot g(x)\big)}$ exists, then both $\displaystyle \lim_{x \to a} {f(x)}$ and $\displaystyle \lim_{x \to a} {g(x)}$ always exist.

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\text{True}

\text{False}

2 solutions

Rishabh Deep Singh
Jan 27, 2017

Take f(x) = x and g(x) = 1/x.

Tapas Mazumdar
Sep 21, 2016

False.

Consider $f(x) = \dfrac{\left| x-a \right|}{x-a}$ and $g(x) = \dfrac{x-a}{\left| x-a \right|}$ .

Then $\displaystyle \lim_{x \to a} {\big( f(x)\cdot g(x) \big)}$ exists, but both $\displaystyle \lim_{x \to a} {f(x)}$ and $\displaystyle \lim_{x \to a} {g(x)}$ do not exist.

Just change to normal curve brackets and you don't need to clarify. I changed it to a bigger pair using \ big( and \ big). Actually when you are using floor function then you clarify and ordinary brackets no need to clarify. Also floor function is $\lfloor \cdot \rfloor$ not brackets.

Chew-Seong Cheong - 4 years, 8 months ago

I know, but in our country the convention is square brackets and it is stated that it represents G.I.F. So, I thought many may get confused whether to treat it as G.I.F. or simple square brackets. The big parentheses are good. Thanks for the tip.

Tapas Mazumdar - 4 years, 8 months ago

They are used for convenience only. Not the official one.

Chew-Seong Cheong - 4 years, 8 months ago

Let's do some calculus! (14)

For more problems on calculus, click here .

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