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Question Number 119253 by bemath last updated on 23/Oct/20

G i v e n t h r e e f u n c t i o n f (x), g (x) a n d h (x) .

w h e r e f (x) = x^{2} + x - 2 a n d h (x) = x^{2} + 2 x - 1 .

I f \frac{f (x)}{x + 3} ⩽ \frac{g (x)}{x + 3} ⩽ \frac{h (x)}{x + 3}, t h e n t h e v a l u e o f

\lim_{x \to - 1} g (x) = ?

Answered by benjo_mathlover last updated on 23/Oct/20

f (x) = (x + 2) (x - 1)

h (x) = {(x + 1)}^{2} - 2

\lim_{x \to - 1} \frac{f (x)}{x + 3} ⩽ \lim_{x \to - 1} \frac{g (x)}{x + 3} ⩽ \lim_{x \to - 1} \frac{h (x)}{x + 3}

- \frac{1}{2} \lim_{x \to - 1} f (x) ⩽ - \frac{1}{2} \lim_{x \to - 1} g (x) ⩽ - \frac{1}{2} \lim_{x \to - 1} h (x)

\Leftrightarrow \lim_{x \to - 1} h (x) ⩽ \lim_{x \to - 1} g (x) ⩽ \lim_{x \to - 1} f (x)

w h e r e \lim_{x \to - 1} h (x) = - 2 \land \lim_{x \to - 1} f (x) = - 2

s o w e g e t \lim_{x \to - 1} g (x) = - 2

Answered by 1549442205PVT last updated on 23/Oct/20

when x \to - 1 then x + 3 > 0 . Hence

\frac{f (x)}{x + 3} ⩽ \frac{g (x)}{x + 3} ⩽ \frac{h (x)}{x + 3}

\Leftrightarrow f (x) ⩽ g (x) ⩽ h (x) . We infer that

\underset{\to - 1}{\lim f} (x) ⩽ \underset{x \to - 1}{\lim g} (x) ⩽ \underset{x \to - 1}{\lim h} (x)

\Leftrightarrow \lim_{x \to - 1} (x^{2} + x - 2) ⩽ \underset{x \to - 1}{\lim g} (x) ⩽ \lim_{x \to - 1} (x^{2} + 2 x - 1)

\Leftrightarrow - 2 ⩽ \underset{x \to - 1}{\lim g} (x) ⩽ - 2

\Rightarrow \underset{x \to - 1}{\lim g} (x) = - 2