Then, as ##b## goes to 0, can you find the limit of each factor separately?

[econmajor]

Homework Statement

a. Compute the limit for f(x) as b goes to 0

Homework Equations

$$f(x) = \frac{(a+bx)^{1-1/b}}{b-1}$$
$a \in R$, $b\in R$, $x\in R$

The Attempt at a Solution

$a+bx$ goes to $a$
$1/b$ goes to $\infty$ so $1-1/b$ goes to $-\infty$
$(a+bx)^{1-1/b}$ then goes to 0
$b-1$ goes to -1

I have no ideas how to combine and conclude. Which mathematical property/rule can I use?

 

[fresh_42]

Homework Statement

a. Compute the limit for f(x) as b goes to 0

Homework Equations

$$f(x) = \frac{(a+bx)^{1-1/b}}{b-1}$$
$a \in R$, $b\in R$, $x\in R$

The Attempt at a Solution

$a+bx$ goes to $a$
$1/b$ goes to $\infty$ so $1-1/b$ goes to $-\infty$
$(a+bx)^{1-1/b}$ then goes to 0
$b-1$ goes to -1

I have no ideas how to combine and conclude. Which mathematical property/rule can I use?

You could simply combine your stuff and find a nominator zero and a denominator minus one, a quotient which can be calculated. However, there is a question: What happens in case $a=1\,$? And is there a different behavior for numbers $a<1$ and $a>1\,$?

Nothing of it is actually a proof, so do you have to prove your result? Or do you just have to calculate the limit by given formulas, in which case I'd like to know which ones do you have?

 

[Mark44]

Homework Statement

a. Compute the limit for f(x) as b goes to 0

Homework Equations

$$f(x) = \frac{(a+bx)^{1-1/b}}{b-1}$$
$a \in R$, $b\in R$, $x\in R$

The Attempt at a Solution

$a+bx$ goes to $a$
$1/b$ goes to $\infty$ so $1-1/b$ goes to $-\infty$
$(a+bx)^{1-1/b}$ then goes to 0
$b-1$ goes to -1

I have no ideas how to combine and conclude. Which mathematical property/rule can I use?

You can't work with the quantities involved in isolation. For example, consider $\lim_{x \to 0} (1 + x)^{1/x}$. One might think this limit is 1, since 1 + x is approaching 1, and 1/x is getting larger and larger. Thinking that 1 to any power is still 1, one might conclude that the limit in my example is 1. However, the correct value is the number e.
I don't have any good advice on your limit, but I'll see if I can come up with something.

 

[vela]

Try rewriting $f(x)$ as
$$f(x) = \frac{a+bx}{b-1}\cdot \frac{1}{(a+bx)^{1/b}}.$$