Question

Limits help again.

Answer 1

http://i.imgur.com/HUaZ7AM.png

Answer 2

am i understanding correctly?

if p=q=r =1

then lim x = infinity
which is a non zero number
but that isnt one of the cases

Answer 3

I am not sure even if the limit exits for other value of p, q ,r other than 0.

Answer 4

it does exist surely

Answer 5

can you give me one set of non zero values of p, q, r? maybe then it would be easier to generalize it.

Answer 6

if x->0 then it's easy case, here n->infty, the problem is lim n->infty sin(x) does not exist. i can't say this is even indeterminate case. since x^p->infty for p>0 and x^p->0 for p<0.

the other part sin^q(x)/sin(x^r) <-- this isn't the indeterminate case ... we know that both numerator and denominator does not exist.

if the above limit were to exist, you need something like sin^q(x)/sin(x^r)~x^(-p)
this is the hard part.

Answer 7

I tried couple of values for q,r on mathematica, it concludes [-infty, +infty] which says the limit does not exist.

Answer 8

for p=q=r=0, the trivial case, limit is csc(1)

Answer 9

\[\Large \lim_{x \rightarrow 2} \frac{(\cos \alpha)^x +(\sin \alpha)^x -1}{x-2}\]
can you hint this one then? :|

Answer 10

this one is L'hopital

Answer 11

i know..i have options
infinity
cos squared alpha+sine squared alpha
0
none of these
is it none of these?
since im getting logs n stuff.

Answer 12

now i come to think of it, the limit does not exist ... possibly.
consider two cases, x->2+ and x->2-

Answer 13

\[\Large (\cos \alpha)^x \times (\log \cos \alpha) + (\sin \alpha)^x \times (\log \sin \alpha)\]
this is the differential^

Answer 14

I thought for a while that it can be zero because log cos alpha and log sine alpha both will tend to 0 nearly because cos and sin can yield max value 1 and therefore the log term will tend to 0,but that is just a thought..

Answer 15

no ... it could be negative, or i can be complex valued. it cannot be zero, because cos and sin are never zero at same time.
does not exist is possibly answer.

Answer 16

LHL =\= RHL

Answer 17

hmmm,none of these again D:

Answer 18

the top one is always negative since sin^2(alpha) + cos^2(alpha) <= 1 (both cases, when x->2+ or x->2-)
the bottom one x-2, it can be both +ve or -ve.

although one sided limit exist.
\[\Large \pm (\cos \alpha)^2 \times (\log \cos \alpha) + (\sin \alpha)^2 \times (\log \sin \alpha) \]

Answer 19

ah i get it :O

Answer 20

i wonder if x->0 instead of x->infinity is the case in your original limit. i would like to see if it exists.

Answer 21

ill let u know the solution tomorrow of the original question

Answer 22

ok see you then