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.

Hi Arnab,

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.

Hope it helps.

All the best.

Regards,

Ashwin (IIT MadraS).

using L Hospital rule its coming 1/2.

Hi Arnab,

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,

Ashwin (IIT MadraS).

extremely sorry , x tends to pi/4, not zero.

Hi Arnab,

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.

All the best.

Regards,

Ashwin (IIT Madras).