Limit it

Calculus Level 4

$\large \lim_{x \to 0}\left(\left \lfloor \frac{100x}{\sin x}\right \rfloor +\left \lfloor \frac{99\sin x}{x}\right \rfloor \right) = \, ?$

Notation : $\lfloor \cdot \rfloor$ denotes the floor function .

198 197 0 199 200

2 solutions

Sparsh Sarode
May 27, 2016

$\displaystyle \lim_{x\rightarrow 0}\dfrac{\sin x}{x}=0.999...$

$\displaystyle \lim_{x\rightarrow 0} \dfrac{x}{\sin x}=1.0000...$

Therefore, $\displaystyle \lim_{x\rightarrow 0} \left( \left[\dfrac{100x}{\sin x} \right]+ \left[\dfrac{99 \sin x}{x} \right] \right) =100+98=198$

But the GIF of 0.999.... = 1

Sabhrant Sachan - 5 years ago

GIF is greatest integer function or floor function... Ceiling function of 0.999=1

Sparsh Sarode - 5 years ago

That was one of the problems I faced when I first learnt floor function and ceiling functions. It was disturbing.

Ashish Menon - 5 years ago

Exactly, same here

Sparsh Sarode - 5 years ago

Simple and easy

Ashish Menon - 5 years ago

Yep, thanks

Sparsh Sarode - 5 years ago

Isn't the limit of sinx/x as x tends to 0 = 1, You can prove that using sandwich theorem or just using woldfram aplha to compute for us. (https://www.wolframalpha.com/input/?i=limit+of+sinx%2Fx+as+x+goes+to+0 ) you can also look here. I personally think the answer should be 199 but since my knowledge is minimum, please teach me.

Tenzin Jampa - 3 years ago

The answer is tending to 1 but since $\sin x < x \ \forall x \in \mathbb{R} \$ the value is always less than 1

Ashish Menon - 3 years ago

Takahiro Waki
May 27, 2018

Think $\lim_{1+0}, \lim_{1-0}$ .

Limit it

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