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lim x tends to 0 (root[2]cosx-1)/(cotx-1)
a) 0
b)1/2
c) 1
d) none of these.
(c)
Hi Arnab,
(√2cosx-1)/(cotx-1) = (√2cosx-1)sinx/(cosx-sinx).
Now this clearly tends to 0, as x tends to 0.
Hence Option (A).
Hope it helps.
Regards,
Ashwin (IIT MadraS).
Answer is 1/2.
1/2 cannot be the answer. It is wrong (must be a printing error in the source from where you got it).
You can clearly see that (√2cosx-1)sinx/(cosx-sinx), will tend to (√2-1)*0/(1-0) = 0, as x tends to 0.
Hence 0 is the answer my dear.
You can check the working yourself.
All the best.
using L Hospital rule its coming 1/2.
You should be careful while applying L-Hospital Rule.
It is applicable on limtis of the form (0/0) or (∞/∞).
This limit is not of that form.
Hence L-Hospital Rule cannot be applied to this limit.
Best Regards,
extremely sorry , x tends to pi/4, not zero.
Never mind.
In that case, you apply L-Hospital rule and get the answer.
But yeah, the question could be x tending to 0 as well. In which case it is 0.
Ashwin (IIT Madras).
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