Right & Left Hand Limits of (sinx)^tanx: Indeterminate Form 0^0

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SUMMARY

The right-hand limit (RHL) of the function (sinx)^tanx as x approaches 0 is 1, while the left-hand limit (LHL) does not exist due to the function being undefined for negative values of x. This conclusion is drawn from analyzing the behavior of the function near the indeterminate form 0^0. The graph of the function confirms that as x approaches 0 from the right, the output approaches 1, while approaching from the left leads to undefined behavior.

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  • Basic knowledge of trigonometric functions, particularly sin(x)
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erisedk
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What is the right hand and left hand limit of (sinx)^tanx?
I know this is an indeterminate form 0^0 but what are the RHL and LHL because though I intuitively know that 0^0 is indeterminate, I don't understand what the right hand and left hand limits are?
 
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As I implement!

$$\lim_{x\rightarrow 0}sinx^{tanx}$$

look at the graph of the function

http://www.wolframalpha.com/input/?i=sin x^{tan x}

And this is intuitively reasonable,
Let us assume a very small value approaching to zero for theta [in radians] , say 0.00000001, then the value will be very close to 1
This is the limit from the right .
And the limit from the lift doesn't exist since you x^x is not always valid for negative numbers..,Hope that is right and what you are looking for ,,
:)
 
Last edited:
It doesn't make sense to talk about "limits" at all without saying what x itself if going to. I presume here you mean "limit as x goes to 0". The "left hand limit" would be as x approaches 0 "from the left" on the number line- that is, x is always negative. The "right hand limit" would be as x approaches 0 "from the right on the number line- x is always positive.

You should be able to see from your graph that if x is approaching 0 "from the right" or "from above", then f(x) goes to 1 while it cannot approach 0 "from the left" or "from below" because, as Maged Saeed said, sin(x)tan(x) is not defined there.
 
Ok thanks :D
 

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