Question

Lim-0(1 + 3x)/x =?????​

Answer 1

Answer:

not exist

Step-by-step explanation:

$\lim\limits_{x\to0}\dfrac{1+3x}{x}\\\\\lim\limits_{x\to0^+}\dfrac{1+3x}{x}=\dfrac{1+(3)(0)}{(+)}=\dfrac{1}{(+)}=+\infty\\\\\lim\limits_{x\to0^-}\dfrac{1+3x}{x}=\dfrac{1+(3)(0)}{(-)}=\dfrac{1}{(-)}=-\infty$

Answer 2

Answer:

See below.

Step-by-step explanation:

The two sided limit does not exist.

As x approaches 0 from negative values the limit is -∞.

As x approaches 0 from positive values the limit is ∞.