Finding the Limit of a Complex Expression

[manasi bandhaokar]

Homework Statement

limit (5/(2+(9+x)^(0.5))^(cosecx)
x-->0
attempt:
tried applying lim (1+x)^(1/x) = e.
x->0
couldn't get anywhere.

 

[physics user1]

Try direct substitution

 

[manasi bandhaokar]

Try direct substitution

wouldn't work.there's a 'cosecx' up there which goes to infinity at x = 0

 

[physics user1]

wouldn't work.there's a 'cosecx' up there which goes to infinity at x = 0

Ahhh you're right, sorry

 

[Mastermind01]

Is it permissible to use L'hopital's rule?

 

[manasi bandhaokar]

Is it permissible to use L'hopital's rule?

yep.but i would also like to know how to solve it the way i was trying to.i thought abt l'hospital but couldn't figure out how to apply it.

 

[Mastermind01]

yep.but i would also like to know how to solve it the way i was trying to.i thought abt l'hospital but couldn't figure out how to apply it.

You have $\lim_{x\rightarrow 0} ({\frac{5}{2+\sqrt{9+x}}})^{cosec (x)}$

The key idea here is to take the logarithm. Since the logarithm is a continuous function log of the limit is the limit of the log.

So the $e^{\lim_{x\rightarrow 0} \ln({\frac{5}{2+\sqrt{9+x}}})^{cosec (x)}}$

This reduces to $e^{\lim_{x\rightarrow 0}\frac{\ln({\frac{5}{2+\sqrt{9+x}})}}{sin (x)}}$

You can apply l'hopital or use other methods to evaluate the top portion.

P.S - Next time while asking homework questions, use the template.

 

[haruspex]

You can clearly replace the cosec x with 1/x immediately.
If you then transform the $\frac 5{2+\sqrt{9+x}}$ into the form 1+ some fraction, you can substitute x=0 immediately in the denominator of that fraction, and expand the numerator with the binomial theorem.

 

[manasi bandhaokar]

Thanks all for the Help.