# Thread: Limitations!

1. ## Limitations!

Hello Everyone! Have here some questions from my maths exercise
Question 1:
lim
x->0 4

It states the limit x approaching 0 however the function f(x) = 4, a single digit!

I have no idea how to do this one..

Second question:

lim f(x)
x-> 3

f(x) = { 2x^2 - 1, x less than or equal to, 3
{ 3x^2 - (30/x), x> 3

How do I do this?

2. Originally Posted by Ace888 Hello Everyone! Have here some questions from my maths exercise
Question 1:
lim
x->0 4

It states the limit x approaching 0 however the function f(x) = 4, a single digit!

I have no idea how to do this one..

Second question:

lim f(x)
x-> 3

f(x) = { 2x^2 - 1, x less than or equal to, 3
{ 3x^2 - (30/x), x> 3

How do I do this?
To #1: There aren't any restrictions therefore the limit is 4.

to #2: There aren't any restrictions either. So you only have to plug in the x-value the function is approaching. The limit is 17 from both sides. Thus the function is continuous but not differentiable at x = 3.

3. Argh I see
When you say a limit does not have any restriction, does that mean that the line or curve goes to infinity? (i.e, 2x +1)

also, for the 2nd question,
the answer is
lim
x-> 3 f(x) = 3

or

Lim
x->3 f(x) does not exist

Another problem thats not related to limits.. but help me please

I dont know how to differentiate Ln (x^4 + (1/x))

thank you so much

4. Originally Posted by Ace888 Argh I see
When you say a limit does not have any restriction, does that mean that the line or curve goes to infinity? (i.e, 2x +1) sorry, that wasn't meant by me. I wanted to point out that there are no restrictions concerning the function itself as division by zero or square-root of negative numbers or logarithm of negative numbers etc.

also, for the 2nd question,
the answer is
lim
x-> 3 f(x) = 3

or

Lim
x->3 f(x) does not exist
I don't know how you got this result ... ? $\displaystyle f(x)=\left\{ \begin{array}{l} 2x^2 - 1,\ x \leq 3 \\ 3x^2 - \dfrac{30}x,\ x> 3 \end{array}\right.$ ...... $\displaystyle \implies$ ...... $\displaystyle \lim_{x \to 3}(f(x))= \left\{ \begin{array}{l} \lim_{x \to 3^-}(2x^2 - 1) = 17 \\ \\ \lim_{x \to 3^+}\left(3x^2 - \frac{30}x \right) = 27-10=17 \end{array}\right.$

(I don't know why the center subscript doesn't work in an array. $\displaystyle \lim_{x \to 3^-}$ means x is approaching 3 from below (or left).

Another problem thats not related to limits.. but help me please

I dont know how to differentiate Ln (x^4 + (1/x))

thank you so much
1. If you have a new question please start a new thread. Otherwise you risk that nobody will notice that you need further support and assistance.

2.
$\displaystyle l(x)= \ln\left(x^4 + \frac1x \right)$ is a logarithm function whose derivation is:

$\displaystyle f(x)=\ln(x)~\implies~f'(c)=\frac1x$

Since you don't have a single x as argument of l you have to use the chain-rule:

$\displaystyle l'(x)=\dfrac1{x^4 + \frac1x } \cdot \dfrac{d\left(x^4 + \frac1x \right)}{dx}$

Simplify!

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