Finding f(x) Using Limit and Algebraic Manipulation

[lendav_rott]

Whoah, I'm interested in this one too now.

If lim x->1 g(x) = c what do you mean by "what g(x) would simplify things" in the product of the 2 limits?

 

[johnqwertyful]

Whoah, I'm interested in this one too now.

If lim x->1 g(x) = c what do you mean by "what g(x) would simplify things" in the product of the 2 limits?

You CREATE a new limit to multiply with the limit you're given. What limit would simplify things?

 

[johnqwertyful]

Just create any g(x), find the limit, see if it simplifies things.

Try g(x)=cos(x). Try g(x)=x^2. Try g(x)=sqrt(x).
See what happens. Does it simplify things?

 

[Curious3141]

Why make the problem so complicated?

The ONLY way for the limit to exist is for the ratio to tend to the indeterminate form $\frac{0}{0}$ as $x \to 1$.

That happens when $\lim_{x \to 1} (f(x) - 8) = 0$, or simply $\lim_{x \to 1} f(x) = 8$.

That's all that's needed.

The RHS doesn't influence that answer at all. What it does is influence the limit of $f'(x)$ (the first derivative) as $x \to 1$. By L' Hopital's Rule, differentiating the numerator and denominator separately gives the result $\lim_{x \to 1} f'(x) = 10$. But this part is just for interest, and not needed for the original answer.

 

[johnqwertyful]

That's another way to do it, although I'm not sure if he cover l'hospital's.I would comment that you don't know f is differentiable; but if it is differentiable, your last paragraph is cool.

 

[Curious3141]

That's another way to do it, although I'm not sure if he cover l'hospital's.


I would comment that you don't know f is differentiable; but if it is differentiable, your last paragraph is cool.

There is no need to invoke L' Hopital's for the answer - that's just algebra.

I only used L' Hopital's Rule for the latter part ("for interest"), which is where the RHS would come into play. I should've stated that f(x) has to be differentiable around the neighbourhood of x = 1 for this to be valid.