Question-94764

Question Number 94764 by Hamida last updated on 20/May/20

Answered by prakash jain last updated on 20/May/20

1 a (i)

\lim_{x \to 3} = 1 (top point of semicircle)

1 a (ii)

\lim_{x \to - 1^{-}} = 0

1 a (iii)

\lim_{x \to 0^{+}} = 0

(iv)

function is not continous at f (0)

\lim_{x \to 0^{-}} f (x) = 0 = \lim_{x \to 0^{+}} f (x)

However f (0) = 2

(v)

\lim_{x \to 3^{-}} f (3) = 1 = \lim_{x \to 3^{+}} f (x)

To makd function continuous

f (3) value should be changed to 1 .

Commented by prakash jain last updated on 20/May/20

\lim_{x \to 0} \frac{\sqrt{2 x^{2} + 4} - 6}{x^{2} - 2 x - 8}

\lim_{x \to 0} \frac{\sqrt{2 x^{2} + 4} - 6}{x^{2} - 2 x - 8} \times \frac{\sqrt{2 x^{2} + 4} + 6}{\sqrt{2 x^{2} + 4} + 6}

\lim_{x \to 0} \frac{2 x^{2} + 4 - 36}{(x - 4) (x + 2)} \times \frac{1}{\sqrt{2 x^{2} + 4} + 6}

\lim_{x \to 0} \frac{2 x^{2} - 32}{(x - 4) (x + 2)} \times \frac{1}{\sqrt{2 x^{2} + 4} + 6}

\lim_{x \to 0} \frac{2 (x + 4) (x - 4)}{(x - 4) (x + 2)} \times \frac{1}{\sqrt{2 x^{2} + 4} + 6}

\lim_{x \to 0} \frac{2 (x + 4)}{(x + 2)} \times \frac{1}{\sqrt{2 x^{2} + 4} + 6}

p u t x = 4

= \frac{2 \times 8}{6} \times \frac{1}{12} = \frac{2}{9}

Commented by student work last updated on 21/May/20

ans : \frac{1}{2}

Answered by mathmax by abdo last updated on 21/May/20

let f (x) = \frac{\sqrt{{2 x}^{2} + 4} - 6}{x^{2} - 2 x - 8} we have x^{2} - 2 x - 8 = x^{2} - 16 - 2 x + 8

= (x - 4) (x + 4) - 2 (x - 4) = (x - 4) (x + 4 - 2) = (x - 4) (x + 2)

\lim_{x \to 4} f (x) = \lim_{x \to 4} \frac{\frac{{2 x}^{2} + 4 - 36}{\sqrt{{2 x}^{2} + 4} + 6}}{(x - 4) (x + 2)}

= \lim_{x \to 4} \frac{{2 x}^{2} - 32}{(x - 4) (x + 2) (\sqrt{{2 x}^{2} + 4} + 6)}

= \lim_{x \to 4} \frac{2 (x - 4) (x + 4)}{(x - 4) (x + 2) (\sqrt{{2 x}^{2} + 4} + 6)}

= \lim_{x \to 4} \frac{2 x + 8}{(x + 2) (\sqrt{{2 x}^{2} + 4} + 6)} = \frac{16}{4 (12)} = \frac{4 \times 4}{4 \times 3 \times 4} = \frac{1}{3}

Commented by prakash jain last updated on 21/May/20

= \lim_{x \to 4} \frac{2 x + 8}{(x + 2) (\sqrt{{2 x}^{2} + 4} + 6)} = \frac{16}{4 (12)} = \frac{4 \times 4}{4 \times 3 \times 4} = \frac{1}{3}

Calculation mistake shown in red

x + 2 = 6

so it should be = \frac{16}{6 \times 12} = \frac{2}{9}