39xx + 1 = x + 1 - 1x + 1 = 1 - 1x + 1
Limx→∞[ (1 - 1(x + 1))-(x + 1)]-xx + 1
= eLimx→∞(-xx + 1)
= e-1?
In limit we often come across terms like 0∞,∞0etc etc.
Now i give some terms .please tell whether they are determinant or determinant and their values if determinant.
1.0∞=(my ans 0)
2.∞0=(my ans 1)
3.∞∞=(my ans ∞)
4.1∞=(my ans 1)
if any other forms please post and lemme know.
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31 Answers
39
1it 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"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]
6hw?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.
6hey dnt mind.......jst give an example....dat wud make thngs clear.........:)i tried bt m gettin stuck....
39Kalyan I used L'Hospital rule in the evaluation of the limit in the power..in post #17
Directly.
6actually 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.......
6can nyone use l'hospital here?in d previous question?
another doubt....limm→∞(cosx/m)m.....
11Ans) 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
6tush bhaiya dont kno d ans.pls post ur working
39All these are indeterminate...if you remember L'Hospital Rule, it can be applied in these very situations.
Even 00 is indeterminate.
11"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
1put it as y=lim
n take Log on both sides............. it will come to 0/0 or ∞/∞ form
39The forms Tushar has mentioned...you can apply L'Hospital on them.
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6ok then.....if such form cums hw to convert into 0/0 or ∞/∞ so dat i can apply l'hospital......
11INDETEMRIANTE FORMS - 00 , ∞∞, 0 x ∞ , ∞ - ∞ , 1 ∞ , 0 0 , ∞ 0,∞ x ∞ , etc
39Because of the very reason I mentioned. You can't apply exponent laws on ∞.
a0 = 1 is an exponent law.
6yea 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?
39Actually, 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"
6forgive 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 .
39"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.