IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

neelesh_tripathi

lim cotx ^sin2x

^{while x tends to 0}

allamraju

I think the ans is 1.Plz tell me whether i am correct,I will explain.

neelesh_tripathi

yes but how????

riku

nice...

sine function follows similar nature as zero,, means,, while argument that is x in sin2x,, tends to 0+ sin2x tends to 0+ and same for 0-.....

On other hand cot x yields infinity ,,, but but,, it gives +infinity at 0+ and -infinity at 0-....refer graph,, ncert, for cotx..
thus limit does not exist!!

samjhe.. toh rate karne mein sharmana kya?/

neelesh_tripathi

the answer in op malhotra is 1 now plz explain

neelesh_tripathi

the answer in op malhotra is 1 now plz explain

neelesh_tripathi

the answer in op malhotra is 1 now plz explain

riku

ohk.. but i doubt such case...

if it would have been ${|cotx|}^{sin2x}$

but yes we can consider such case also..

since its tending to infinity,, it will be a number.. and for any number "n",,

${n}^{0}={n}^{1-1}=\frac{{n}^{1}}{{n}^{1}}=1$

thus limit becomes tending to 1..

allamraju

My method is this; Take the limit to be k.Then lnk=Lim sin2x.ln(cotx)

Write this as ln(cotx)/1/sin2x.It is now of infinity/infinity form.Apply L'hospital rule once and simplify,it reduces to tan2x which tends to 0 as x tends to 0.lnk=0 implies k=1

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I think the ans is 1.Plz tell me whether i am correct,I will explain.

Write this as ln(cotx)/1/sin2x.It is now of infinity/infinity form.Apply L'hospital rule once and simplify,it reduces to tan2x which tends to 0 as x tends to 0.lnk=0 implies k=1