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Pritish Chakraborty
·2010-05-24 11:46:40
xx + 1 = x + 1 - 1x + 1 = 1 - 1x + 1
Limx→∞[ (1 - 1(x + 1))-(x + 1)]-xx + 1
= eLimx→∞(-xx + 1)
= e-1?
1
terror s
·2010-05-25 04:51:14
it is not indeterminant.......
in limits we say ot indeterminant bcz the value always tend to and not the exact value...
thats why.
for eg: ∞°=1 only when the power is 0 and not when it tends to ..
11
Tush Watts
·2010-05-24 12:41:11
"can nyone use l'hospital here?in d previous question?"
Ques) lim y→0 (11+y) 1/y
Ans) Approach by L'H rule
Take log both sides , we get
log L =lim y→0 log (11+y)y
now since it's indeterminate form, u can use L'H rule, and u will get
log L = -1 , so L = e -1
Thus proved. [1] [3]
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Tush Watts
·2010-05-24 12:33:32
Ans) ok it's coming to be 0 ????
6
Kalyan IIT-K Beware I'm coming
·2010-05-24 12:21:59
ohhhhhhhh m really sorry its m nt x.........
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Tush Watts
·2010-05-24 12:18:58
@kalyan , is it x→∞ or m→∞ ????
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Kalyan IIT-K Beware I'm coming
·2010-05-24 12:10:14
hw?i got d same....bt using a logic which was wrong.......
cos(0)∞ is wat we get puttin limit i.e. 1∞.i assumed it as one.
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Tush Watts
·2010-05-24 12:06:37
Ques) limx→∞ (cosx/m)m
Ans) 1 ???????
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Kalyan IIT-K Beware I'm coming
·2010-05-24 12:06:16
hey dnt mind.......jst give an example....dat wud make thngs clear.........:)i tried bt m gettin stuck....
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Tush Watts
·2010-05-24 12:04:20
To convert into 0/0 or ∞/∞ form, take log both sides
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Pritish Chakraborty
·2010-05-24 11:58:38
Kalyan I used L'Hospital rule in the evaluation of the limit in the power..in post #17
Directly.
6
Kalyan IIT-K Beware I'm coming
·2010-05-24 11:54:36
actually i was doing a serious mistake by assuming 1.0∞=(my ans 0)
2.∞0=(my ans 1)
3.∞∞=(my ans ∞)
4.1∞=(my ans 1)
nd i was solving sums.......
6
Kalyan IIT-K Beware I'm coming
·2010-05-24 11:53:27
can nyone use l'hospital here?in d previous question?
another doubt....limm→∞(cosx/m)m.....
11
Tush Watts
·2010-05-24 11:52:50
Ans) lim x→∞ (xx+1 ) x
Put 1/x = y, so y → 0
= lim y→0 (11+y) 1/y
= e lim y→0 (11+y -1 ) 1y
= e lim y→0 (-y1+y ) 1y
= e -1
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Kalyan IIT-K Beware I'm coming
·2010-05-24 11:46:57
tush bhaiya dont kno d ans.pls post ur working
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Pritish Chakraborty
·2010-05-24 11:10:29
All these are indeterminate...if you remember L'Hospital Rule, it can be applied in these very situations.
Even 00 is indeterminate.
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Tush Watts
·2010-05-24 11:38:36
"ok then.....if such form cums hw to convert into 0/0 or ∞/∞ so dat i can apply l'hospital......"
By using log / exponential function
1
rickde
·2010-05-24 11:38:35
put it as y=lim
n take Log on both sides............. it will come to 0/0 or ∞/∞ form
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Pritish Chakraborty
·2010-05-24 11:37:11
The forms Tushar has mentioned...you can apply L'Hospital on them.
-EDITED-
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Kalyan IIT-K Beware I'm coming
·2010-05-24 11:34:14
ok then.....if such form cums hw to convert into 0/0 or ∞/∞ so dat i can apply l'hospital......
11
Tush Watts
·2010-05-24 11:33:40
INDETEMRIANTE FORMS - 00 , ∞∞, 0 x ∞ , ∞ - ∞ , 1 ∞ , 0 0 , ∞ 0,∞ x ∞ , etc
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Pritish Chakraborty
·2010-05-24 11:32:16
Because of the very reason I mentioned. You can't apply exponent laws on ∞.
a0 = 1 is an exponent law.
6
Kalyan IIT-K Beware I'm coming
·2010-05-24 11:25:47
yea dats wat i was saying.dat since x is ∞ nd nt finite so it cant go like dat.thus my pt was ∞0is 1.y cant it be?
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Pritish Chakraborty
·2010-05-24 11:22:02
Actually, you can't do that. ∞ is not a number on the real number line. Any laws which pertain to the real number line(such as exponent transformation which you have done) is not applicable to ∞.
In short, as our dear old DOS would say....
"Bad command or filename"
6
Kalyan IIT-K Beware I'm coming
·2010-05-24 11:20:27
forgive me but is still think dat my answers r right.∞0 can be written as ∞1-1 i .e.∞/∞ bt xa-b =xa/xb is only valid when x is finite .
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Pritish Chakraborty
·2010-05-24 11:15:06
"The most common example of an indeterminate form is 0/0. As x approaches 0, the ratios x/x3, x/x, and x2/x go to infinity, 1, and 0 respectively. In each case, however, if the limits of the numerator and denominator are evaluated and plugged into the division operation, the resulting expression is 0/0. So (roughly speaking) 0/0 can be 0 or it can be infinity and, in fact, it is possible to construct similar examples converging to any particular value. That is why the expression 0/0 is indeterminate." - Wiki
Now you can understand why the others are indeterminate too.
However, ∞/0 and 0/∞ are not indeterminate but are undefined. Any limit which produces these forms converges to zero or diverges, as per Wiki.