Limit

Question

The limit x->0  sin[2x/3]/x is

Answer 1

$\lim_{x \to 0} \frac{sin\frac{2x}{3}}{x}$

Putting x = 0 on num and deno we get 0/0 form.

Applying L'Hopital's Rule:-

Differentiate the num and deno separately.

we get:-

$\lim_{x \to 0} \frac{\frac{2}{3}cos\frac{2x}{3}}{1}$

Put x = 0

You will get

$\frac{2}{3}$

Comments

We can directly reduce greatest integer function[2x/3] to 2x/3 is it possible?

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if it is greatest integer function ,,,then limit not exist

f(0⁺) = 0

f(0^-) = -inf

f(0⁺) != f(0^-)

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@saumya mishra

@shubhanshu has done without considering [ ] as greatest integer function, you should mention it in qsn..

if it is greatest integer function than limit x->0  sin[2x/3]/x is not defined, since

limit x->0^-  sin[2x/3]/x = -infinity

limit x->0⁺  sin[2x/3]/x = 0

LHL not equal to RHL, therefore limit does not exist

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In made easy solution book they have also ignored the greatest integer function and solved in the same way as shubhanshu does.

Answer 2

limit _x->0  sin(2x/3) / x

= limit _x->0  2/3 * {sin(2x/3) / (2x/3)}

=2/3*1 = 2/3

as we know  limit _x->0  sin(x) / x  = 1