Theorem for Limits: Why Is It True?

[Mr Davis 97]

I read in a calculus book that. "Given $\lim_{x \to a}\frac{f(x)}{g(x)} = c(c\neq 0)$, when $\lim_{x \to a}g(x) = 0$, then $\lim_{x \to a}f(x) = 0$. Why is this true?

 

[DEvens]

http://en.wikipedia.org/wiki/L'Hôpital's_rule

 

[Mark44]

I read in a calculus book that. "Given $\lim_{x \to a}\frac{f(x)}{g(x)} = c(c\neq 0)$, when $\lim_{x \to a}g(x) = 0$, then $\lim_{x \to a}f(x) = 0$. Why is this true?

A non-rigorous explanation is that, since $\frac{f(x)}{g(x)} \to c$, where c ≠ 0, then f and g are approximately equal near a. If g approaches zero as x approaches a, then so does f.

 

[PeroK]

I read in a calculus book that. "Given $\lim_{x \to a}\frac{f(x)}{g(x)} = c(c\neq 0)$, when $\lim_{x \to a}g(x) = 0$, then $\lim_{x \to a}f(x) = 0$. Why is this true?

Have you tried finding a counterexample? Usually a good way to see why something is true is to try to show that it's false.

And, why must you have $c \ne 0$?

 

[Mark44]

http://en.wikipedia.org/wiki/L'Hôpital's_rule

I[/PLAIN] don't see how this applies to the OP's question. We already know what the limit is.

 

[mathman]

I read in a calculus book that. "Given $\lim_{x \to a}\frac{f(x)}{g(x)} = c(c\neq 0)$, when $\lim_{x \to a}g(x) = 0$, then $\lim_{x \to a}f(x) = 0$. Why is this true?

Multiply by g(x). Limit for f(x) = c(limit for g(x)) = 0.