Marco asked in Science & MathematicsMathematics · 4 months ago

# Calculus (limits) question?

Given: lim x→0 of f(x)/x = 1

Determine: lim x→0 of f(x)

Here is my attempted solution (what did I do wrong, if anything?)...

Since lim x→0 of f(x)/x = 1, then using quotient rule, this means that:

lim x→0 of f(x)

-------------------- = 1

lim x→0 of x

provided that lim x→0 of x ≠ 0. Also means lim x→0 of f(x) ≠ 0, because that would automatically make the above equation untrue.

Multiplying both sides of the equation by lim x→0 of x, you get:

lim x→0 of f(x) = lim x→0 of x

Which makes sense, both limits are equal. Correct me if I'm wrong, but you cannot plug in 0 for x on the right side, because it was stated earlier that lim x→0 of x ≠ 0. The answer in the back of the book is 0, but that cannot be correct. Any help/ideas?

Relevance
• 4 months ago

It's wonderful that you gave it a shot. Better than the garbage that just seeks answers.

OK, so you claim by the quotient rule that:

lim x->0 f(x)/x = lim x-> f(x)/ lim x->0 x if lim x -> 0 x =/= 0. However, it does, lim x->0 x = 0.

A formal proof is quite simple, let delta = epsilon, then |x - 0| < delta = epsilon. Anyhow, the key point here is that lim x-> 0 x = 0, so the quotient rule fails. I.e., your first equivalence is incorrect. The issue here is that you're confusing the conditional with its converse. The statement is "the quotient rule holds if the denominator doesn't go to zero", it is not "if the quotient rule holds, then the denominator doesn't go to zero" with you assuming that that quotient rule holds in the first place.

Intuitively, since the denominator goes to 0, the numerator would have to go to zero at an equal rate or so in order to have the sequence of approximate values approach 1 as x -> 0.

For proof: for every epsilon > 0 there exists a delta_1 > 0 such that, for |x - 0| < delta we have |f(x)/x - 1| < epsilon.

Let delta = min {epsilon/(epsilon + 1), delta_1}

Then |f(x) - 0| = |x||f(x)/x| = |x||f(x)/x - 1 + 1| <= |x|(|f(x)/x - 1| + 1)

In the delta neighborhood, |x| < delta, i.e. < delta (|f(x)/x - 1| + 1)

Since delta <= delta_1, |f(x)/x - 1| < epsilon, hence

< delta(epsilon + 1) <= epsilon/(epsilon + 1)(epsilon + 1) = epsilon.

I might have screwed up the proof, but the intuition is true as well as why your above approach didn't work.

• Marco4 months agoReport

Thank you!