If-lim-x-a-f-x-g-x-exists-and-lim-x-a-g-x-0-then-show-that-lim-x-a-f-x-0-

Question Number 4363 by Rasheed Soomro last updated on 13/Jan/16

I f \underset{x \to a}{l i m} \frac{f (x)}{g (x)} e x i s t s a n d \underset{x \to a}{l i m} g (x) = 0,

t h e n s h o w t h a t

\underset{x \to a}{l i m} f (x) = 0 .

Answered by Yozzii last updated on 13/Jan/16

L e t l = \lim_{x \to a} \frac{f (x)}{g (x)} a n d h (x) = \frac{f (x)}{g (x)} (g (x) \neq 0) . l e x i s t s i f, f o r e v e r y

ϵ > 0 t h e r e e x i s t s a δ > 0 s u c h t h a t

∣ h (x) - l ∣< ϵ w h e n e v e r 0 <∣ x - a ∣< δ .

∴∣ \frac{f (x)}{g (x)} - \lim_{x \to a} \frac{f (x)}{g (x)} ∣< ϵ

∣ \frac{f (x)}{g (x)} - \frac{\lim_{x \to a} f (x)}{\lim_{x \to a} g (x)} ∣< ϵ

∣ \frac{f (x) \times \lim_{x \to a} g (x) - g (x) \lim_{x \to a} f (x)}{g (x) \lim_{x \to a} g (x)} ∣< ϵ

∴∣ f (x) \lim_{x \to a} g (x) - g (x) \lim_{x \to a} f (x) ∣< ϵ ∣ g (x) \lim_{x \to a} g (x) ∣

S i n c e \lim_{x \to a} g (x) = 0 w e h a v e

∣ 0 - g (x) \lim_{x \to a} f (x) ∣< 0

H e n c e ∣ - g (x) \lim_{x \to a} f (x) ∣< 0 w h i c h i s f a l s e .

(I f i n d e e d \lim_{x \to a} f (x) = 0, t h e r e e x i s t s

s o m e e r r o r i n m y u s e o f t h e ϵ - δ

d e f i n i t i o n o f t h e l i m i t .)

Commented by prakash jain last updated on 13/Jan/16

\lim_{x \to 3} \frac{x^{2} - 9}{x - 3}

Commented by Yozzii last updated on 13/Jan/16

M y e r r o r w a s i n c o r r e c t l y u s i n g t h e

r e s u l t t h a t \lim_{x \to a} \frac{f (x)}{g (x)} = \frac{\lim_{x \to a} f (x)}{\lim_{x \to a} g (x)} g i v e n

t h a t \lim_{x \to a} g (x) \neq 0 .

Answered by RasheedSindhi last updated on 14/Jan/16

A w a y w h i c h {d o e s n}^{'} t u s e ϵ - δ

d e f i n i t i o n .

L e t \underset{x \to a}{l i m} \frac{f (x)}{g (x)} = l

f (x) = \frac{f (x)}{g (x)} \times g (x)

A p p l y i n g l i m i t t o b o t h s i d e s

\underset{x \to a}{l i m f} (x) = \underset{x \to a}{l i m} (\frac{f (x)}{g (x)} \times g (x))

= \underset{x \to a}{l i m} (\frac{f (x)}{g (x)} \times g (x)) = \underset{x \to a}{l i m} \frac{f (x)}{g (x)} \times \underset{x \to a}{l i m} g (x)

= l . 0 = 0