1Nishant if x will tend to zero it will be infinity
but since it is tending to infinity it will be zero
suppose
now method 1,
L=\lim_{x\rightarrow \propto } \frac{1-sinx/x}{1+sinx/x}
since (sinx /x) →0
L=1
now method 2,
using L Hospital rule ( for ∞/∞ form) ,
L=\lim_{x\rightarrow \propto } \frac{1-cosx}{1+cosx}
L=\lim_{x\rightarrow \propto } \frac{2sin^2x/2}{2cos^2x/2}
L=\lim_{x\rightarrow \propto } tan^2x/2
now comparing with previous result, L=1
hence tan2x/2 =1 (x →∞)
hence tanx=±∞ , sinx=±1 , cos x=0
is this correct ???
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5 Answers
21This is what i think.
Using method 2 you got a result that exists but is indeterminate.(this means that it may be an exact value but cannot be determined by just using this one expression)
Using the other method you got the exact limiting value of the same expression.
Hence it is correct to equate the two.
Just a small error tan2x/2→1 and not equal to one.
Hence the following values are the limiting value of the trig. ratios.
But if we just look just into this line
limx→∞ tanx =±∞
it may seem as if it is not true and a bit absurd.
But to make this statement true we must also state that:
let L2=limx→∞ tanx=limx→∞ 2tanx/2
1-tan2x/2
we know L=limx→∞ tan2x/2 .
We can also say that limx→∞ tanx/2=±L (because L=1 )
hence L2=2*±L
1-L
Where L=limx→∞ x+sinx =1
x-sinx
Now we can say L2=2*±L/(1-L)=±∞
Hence L2=limx→∞ tanx=±∞
Now this line makes sense.(If and only if the bold lines are true)
62oh!
the first one sin x/ x is not equal to 1 !!!
here x does not tend to 0 but to infinity
so ur solution 2 is correct.
even for that u should have simply used sandwich theorem to say
(x-1)/(x+1) < L< (x+1)/(x-1)
And x goes to infinity.
So L = 1
341since the limit for f'(x)/g'(x) does not exist L'hospital's rule is not applicable. This in no way implies that the original limit does not exist. You can try this for say (x+ cos x)\x as x→∞. Hid this as I wasnt sure whether you were looking for a discussion or a solution
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