Problem of the Day 4
x → 0 lim x + cos 2 ( x ) x − sin ( x ) = ?
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1 solution
i think you should also show,LHL = RHL because that is the main aim why i added this question...
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Done Akhil Bansal :) :)
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Still incomplete solution,
You should show that . .
&
&
&
{
When
x
<
0
⇒
sin
(
x
)
>
x
,
numerator will be positive
When
x
>
0
⇒
x
>
sin
(
x
)
,
numerator will again be positive
You should numerator is positive because if it any point,numerator becomes negative,limit will not exist.(as denominator is clearly positive for both LHL and RHL)
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@Akhil Bansal – There's no need to do all these. You can do all the operations inside the radical first and show that it's positive at x = 0. Take its squareroot and it's still 0.
What is LHL and RHL?
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LHL= left hand limit and RHL = right hand limit :)