limits

can anyone help me out in finding limits by ε-δ method?? exams mein aa rahaa hai yeh sab... although simple limits hi hain..

8 Answers

62
Lokesh Verma ·

which exams?

106
Asish Mahapatra ·

college ke...

1
ANKIT MAHATO ·

Target iiT mein college ?

62
Lokesh Verma ·

oh asish you are in college//

given epsilon >0, there exists delta >0 such that

f(x)-f(y) < epsilon

for all y such that

|y-x|<delta..

that is all you have to porve..

106
Asish Mahapatra ·

bhaiyya can u give sum simple example like how to find the following limit by that method..
lim(x→2) 2x+3

341
Hari Shankar ·

This is useful when you have conjectured a limit (or its non existence) and want to prove that conjecture.

Here, the limit seems to be 7.

So, we have to prove that for any ε>0, we can exhibit δ(ε) such that for all x such that |x-2|<δ, it is true that |f(x) - 7 |<ε

This is best remembered as the ε-δ challenge. Someone gives you an ε which is usually very very small and you are expected to produce δ corresponding to this ε. That is why we write δ(ε).

So now,

|f(x) - 7| < \epsilon \Rightarrow |2x+3-7| < \epsilon \Rightarrow |x-2| < \frac{\epsilon}{2}

So if we choose \delta = \frac{\epsilon}{2} the stipulated challenge is met.

Hence we have proved that the limit is indeed 7.

This method is then used to establish other methods such as the Sandwich Principle which are more useful to evaluate limits. For example \lim_{x \rightarrow 0} \frac{\sin x}{x}

66
kaymant ·

@ashish,
If you now have got the idea, try to prove that limx→2 x2 = 4.

106
Asish Mahapatra ·

no sir... i cudnt get anything ...because our whole book on limits is completely based on epsilon delta... theres no concept of LHL and RHL... so maybe i need to just forget wats written in the book and read that again.. with cool mind.. maybe i can understand then

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