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  • Evaluate <img alt="\mathrm{\lim _{x \rightarrow 0}\left[\frac{3 \sin x}{x}\right]+\left[\frac{4 x}{\tan x}\right]}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bx%2

Evaluate \mathrm{\lim _{x \rightarrow 0}\left[\frac{3 \sin x}{x}\right]+\left[\frac{4 x}{\tan x}\right]} (where [.] denotes greatest integer function)
 

Option: 1

0


Option: 2

4


Option: 3

6


Option: 4

5


Answers (1)

best_answer

In the proof of \mathrm{\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1}, we had established an inequality for \mathrm{\theta(\theta>0)\sin \theta<\theta<\tan \theta}
Hence \mathrm{\frac{\sin \theta}{\theta}<1}, i.c., we can \mathrm{\operatorname{say} \frac{\sin \theta}{\theta}}approaches 1 from the left-hand side of 1 . Also \mathrm{\frac{\sin \theta}{\theta}} is an even function. So we can say the same thing for \mathrm{\theta<0}. Hence \mathrm{\lim _{\theta \rightarrow r \rightarrow 0}\left[\frac{\sin \theta}{\theta}\right]=0} (where [.] denotes greatest integer function). Similarly it can be established from the right inequality \mathrm{\left[\frac{\tan \theta}{\theta}\right]>1}. Hence, \mathrm{\lim _{\theta \rightarrow 0}\left[\frac{\tan \theta}{\theta}\right]=1} (where [.] denotes greatest integer function). Coming back to our original question we can say.

\mathrm{ \lim _{x \rightarrow 0}\left[\frac{3 \sin x}{x}\right]=2\left(\text { As } \frac{3 \sin x}{x}<3\right) \text { and } \lim _{x \rightarrow 0}\left[\frac{4 x}{\tan x}\right]=3 }

As
\mathrm{ \frac{\tan x}{x}>1 }
\mathrm{ \Rightarrow \quad \frac{x}{\tan x}<1; }
Hence
\mathrm{ \frac{4 x}{\tan x}<4 }

\mathrm{ \therefore \quad \lim _{x \rightarrow 0}\left[\frac{3 \sin x}{x}\right]+\left[\frac{4 x}{\tan x}\right]=5 }

Hence option 4 is correct.
 

Posted by

Irshad Anwar

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