Question

USe Squeeze Theorem to determine
lim x->0 (x+1) cos( ln x^2) / √x^2+2

Answer 1

\[\lim_{x \rightarrow 0} \frac{ (x+1)\cos (\ln x ^{2)} }{ \sqrt{x ^{2} +2} }\]

Answer 2

start with
\[-1 \le \cos(\ln x^2) \le1\]

Answer 3

so

\[\frac{ (x+1) }{ \sqrt{x^2+2 }} \le \frac{ (x+1) }{ \sqrt{x+2} }\cos(\ln x^2) \le -\frac{ (x+1) }{ \sqrt{x+2}}\]
getting this?

Answer 4

now according to squeeze theorem evaluate
\[\lim_{x \rightarrow 0}\frac{ (x+1) }{ \sqrt{x+2} }\]
can u do that?

Answer 5

ok just simply evaluate the limit by setting x=0
\[\lim_{x \rightarrow 0}\frac{ (x+1) }{ \sqrt{x+2} }=\frac{ 0+1 }{ \sqrt{0+2} }=\frac{ 1 }{ \sqrt{2} }\]
and for the left hand side
\[\lim_{x \rightarrow 0}\frac{ -(x+1) }{ \sqrt{x+2} }=\frac{ -1 }{ \sqrt{2} }\]
so u can see \[Left.hand.limit \neq Right.hand.limit\]
so \[LIMIT~DOES~NOT~EXISTS\]

Answer 6

@jibirajeev

Answer 7

let me try. in squeeze theorem we have to get two points one from left and right and our answer must be inbetween. am i right?

Answer 8

Thanks.. got it..:)

Answer 9

yeah thats how we proceed YW!!!!

Answer 10

Thanks... i do understand what you done for me here. if you dont mind can you explain one question with limit does exist.