Limits:(

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November 19th 2008, 08:11 AM
t1i2m3xx

Limits:(

Hey all

This is my first ever post so here goes!!

REALLY stuck on two types of question about limits of functions.

In both cases a proof from first principle is required. In case one I need to prove that if f(x) tends to 5 as x tends to 1 then (f(x))^2 tends to 25 as x tends to 1.

Secondly in another question, I need to show that 2/3+x tends to 0 as x tends to infinity.

I am really stuck as to how to even start these sorts of questions(Crying), any help would really be appreciated.
November 19th 2008, 08:37 AM
james_bond

For the first you have to know the third one from this (but it is good to know all): $\begin{matrix} \lim\limits_{x \to p} & (f(x) + g(x)) & = & \lim\limits_{x \to p} f(x) + \lim\limits_{x \to p} g(x) \\ \lim\limits_{x \to p} & (f(x) - g(x)) & = & \lim\limits_{x \to p} f(x) - \lim\limits_{x \to p} g(x) \\ \lim\limits_{x \to p} & (f(x)\cdot g(x)) & = & \lim\limits_{x \to p} f(x) \cdot \lim\limits_{x \to p} g(x) \\ \lim\limits_{x \to p} & (f(x)/g(x)) & = & {\lim\limits_{x \to p} f(x) / \lim\limits_{x \to p} g(x)} \end{matrix}$
(the last provided that the denominator is non-zero). In each case above, when the limits on the right do not exist, or, in the last case, when the limits in both the numerator and the denominator are zero, nonetheless the limit on the left, called an indeterminate form, may still exist — this depends on the functions $f$ and $g$ .
You can find proof for these (or if you can't I can send you one).

For the second one I suppose you meant $\frac 2{3+x}$ . This is easy just use the definition:
If $f(x)$ is a real function, then the limit of $f$ as $x$ approaches infinity is $L$ , denoted

$\lim_{x \to \infty}f(x) = L$ ,

if and only if for all $\epsilon > 0$ there exists $S > 0$ such that $|f(x) - L| < \epsilon$ whenever $x > S$ .
November 19th 2008, 08:38 AM
Jhevon

Quote:

Originally Posted by t1i2m3xx

Hey all

This is my first ever post so here goes!!

REALLY stuck on two types of question about limits of functions.

In both cases a proof from first principle is required. In case one I need to prove that if f(x) tends to 5 as x tends to 1 then (f(x))^2 tends to 25 as x tends to 1.

$\lim_{x \to 1}f(x) = 5$ means that for every $\epsilon > 0$ , there exists some $\delta > 0$ , such that (x and 1 in the domain of f and) $|x - 1| < \delta$ implies $|f(x) - 5| < \epsilon$ .

You need to show, then, that for any $\epsilon > 0$ , there is some $\delta > 0$ , such that $|x - 1| < \delta$ implies $|(f(x))^2 - 25| < \epsilon$

hint: applying the difference of two squares should help.

Quote:

Secondly in another question, I need to show that 2/3+x tends to 0 as x tends to infinity.

You need to show that given any $A > 0$ , there is some $k \in \text{dom}(f)$ , such that $x \in \text{dom}(f)$ and $x > k$ implies $f(x) > A$ .

Here of course, $f(x) = \frac 2{3 + x}$
November 19th 2008, 08:48 AM
Jhevon

Quote:

Originally Posted by james_bond

For the first you have to know the third one from this (but it is good to know all): $\begin{matrix} \lim\limits_{x \to p} & (f(x) + g(x)) & = & \lim\limits_{x \to p} f(x) + \lim\limits_{x \to p} g(x) \\ \lim\limits_{x \to p} & (f(x) - g(x)) & = & \lim\limits_{x \to p} f(x) - \lim\limits_{x \to p} g(x) \\ \lim\limits_{x \to p} & (f(x)\cdot g(x)) & = & \lim\limits_{x \to p} f(x) \cdot \lim\limits_{x \to p} g(x) \\ \lim\limits_{x \to p} & (f(x)/g(x)) & = & {\lim\limits_{x \to p} f(x) / \lim\limits_{x \to p} g(x)} \end{matrix}$
(the last provided that the denominator is non-zero). In each case above, when the limits on the right do not exist, or, in the last case, when the limits in both the numerator and the denominator are zero, nonetheless the limit on the left, called an indeterminate form, may still exist — this depends on the functions $f$ and $g$ .
You can find proof for these (or if you can't I can send you one).

also, these theorems only hold in general if both limits are finite.

for consider $\lim_{x \to 0} f(x)$ and $\lim_{x \to 0} g(x)$ , where $f(x) = x^2$ and $g(x) = \frac 1{x^2}$

then $\lim_{x \to 0} (fg)(x) \ne ( \lim_{x \to 0} f(x) ) ( \lim_{x \to 0} g(x))$ .

indeed, the left hand side is 1, while the right hand side is indeterminate

i guess this is what you were explaining about the indeterminate form. i believe we only have such a case if either limit (of f or of g or of both) diverges or doesn't exist. if both are finite, i believe the theorems hold. with the exception of the last one where you may have a division by zero, of course