Question

Find if limits exists..

a)lim x tends to 3 .. (2x+ |x-3|)
b) g(x) = x^2+x-6 / | x-2 |

Help me with this.. pls explain modulus in these sums are confusing.

Answer 1

(a)

$Given:\lim_{x \to 3} (2x + |x - 3|)$

$=>\lim_{x \to 3} (2x + x - 3)$

$=>\lim_{x \to 3} (3x - 3)$

= > 3(3) - 3

= > 9 - 3

= > 6.

-----------------------------------------------------------------------------------------------------------------

(b)

$Given:\lim_{x \to 3} \frac{x^2 + x - 6}{|x - 2|}$

$=>\lim_{x \to 3} \frac{x^2 + x - 6}{x - 2}$

$=>\frac{3^2 + 3 - 6}{3 - 2}$

$=>\frac{9 + 3 - 6}{1}$

$=>\frac{6}{1}$

=> 6.

----------------------------------------------------------------------------------------------------------------

Hope it helps!

Answer 2

$see \: the \: above \: attachments \\ you \: get \: the \: answer. \\ \\ hope \: it \: helps...$
Plese Mark As Brainliest