How to solve limits?

**RCola** · December 14th 2010, 05:12 PM

1. Limit [x->1] [(e^x-e)*(x-1)] = e

How to solve ? Derivation (L'Hopital's rule) don't get anything useful - it remains [0/0].

2. Limit [x->0] [Ln[Cos[x]] / x^2] = -1/2

Any idea ?

3. Limit [x->Inf] [(x^3+2x-1)^(1/3) / (x+2)] = 1

How to solve ?

4. [a^b]' = b*Ln[a]*[a]'
Is it right?

**dwsmith** · December 14th 2010, 05:22 PM

$\displaystyle \lim_{x\to 1}\frac{e^x-e}{x-1}=\frac{0}{0}\Rightarrow \lim_{x\to 1}\frac{e^x}{1}=e$
I think you need a division sign not multiplication.

$\displaystyle \lim_{x\to 0}\frac{0}{0}\Rightarrow \lim_{x\to 0}\frac{\frac{-sin(x)}{cos(x)}}{2x}=\frac{0}{0}\Rightarrow\lim_{x \to 0}\frac{-sec^2(x)}{2}=\frac{-1}{2}$
L'Hopital's Rule twice.

$\displaystyle \lim_{x\to \infty}\frac{(x^3)^{1/3}}{x}=\lim_{x\to \infty}\frac{x}{x}=1$
Limits to infinity isolate the highest powered exponents in the numerator and denominator.

What is going on for number 4?

**RCola** · December 14th 2010, 05:32 PM

dwsmith:
How? If we apply L'Hopital's rule = > (-e+x*e^x)/(x-1)^2 => (-e+e)/(1-1)^2 => [0/0] ???

**dwsmith** · December 14th 2010, 05:33 PM

e is constant so the derivative of e is 0 and the same for 1.

The derivative of e^x is e^x and the derivative of x is 1.

**skeeter** · December 14th 2010, 05:36 PM

Originally Posted by RCola

1. Limit [x->1] [(e^x-e)*(x-1)] = e

here is what you originally posted ...

$\displaystyle \lim_{x \to 1} (e^x-e)(x-1)$

this limit is 0 ... what limit expression did you really mean to post?

**dwsmith** · December 14th 2010, 05:37 PM

Originally Posted by skeeter

here is what you originally posted ...

$\displaystyle \lim_{x \to 1} (e^x-e)(x-1)$

this limit is 0 ... what limit expression did you really mean to post?

Division.

**RCola** · December 14th 2010, 05:47 PM

Originally Posted by dwsmith

e is constant so the derivative of e is 0 and the same for 1.

The derivative of e^x is e^x and the derivative of x is 1.

Aren't we suppose to derivate whole function instead of separately derivating numerator and denominator ?

Derivating whole function I've got :

[(e^x-e)/(x-1)^1]' = (e^x)*(x-1)^(-1) + (e^x-e)(-(x-1)^(-2)) = (e^x(x-1) - e^x+e ) / (x-1)^2 = e^x / (x-1)

So : lim[x->1, (e^x)/(x-1)] = e/0 => we aren't allowd to devide by zero => deadlock

Have I somewhere got mistake ?

**dwsmith** · December 14th 2010, 05:48 PM

No, derivative of the numerator and denominator separately.

**RCola** · December 14th 2010, 05:48 PM

Originally Posted by skeeter

here is what you originally posted ...

$\displaystyle \lim_{x \to 1} (e^x-e)(x-1)$

this limit is 0 ... what limit expression did you really mean to post?

Srry, It has to be (e^x-e)/(x-1)

**RCola** · December 14th 2010, 05:51 PM

Originally Posted by dwsmith

No, derivative of the numerator and denominator separately.

Oh, I see, Lim [f(a)/f(b)] = lim [f'(a)/f'(b)]
thx

**Prove It** · December 14th 2010, 05:57 PM

Originally Posted by RCola

Oh, I see, Lim [f(a)/f(b)] = lim [f'(a)/f'(b)]
thx

Only if the function tends to $\displaystyle \frac{0}{0}$ or $\displaystyle \frac{\infty}{\infty}$ .

**RCola** · December 14th 2010, 06:05 PM

Originally Posted by dwsmith

What is going on for number 4?

is it allowed to use this technique to find derivative ?

**dwsmith** · December 14th 2010, 06:10 PM

Derivative with respect to what? a, b, x? Are a and b Real numbers?

**RCola** · December 14th 2010, 06:15 PM

Originally Posted by dwsmith

Derivative with respect to what? a, b, x? Are a and b Real numbers?

with respect to x
a and b is Real numbers

**dwsmith** · December 14th 2010, 06:17 PM

Derivative is 0 then.

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