Question

Lim x approaches 0 (1+2x)3/sinx

Answer 1

Interpreting your expression as

$\dfrac{3(1+2x)}{\sin(x)}$

when $x$ approaches zero, the numerator approaches 3:

$3(1+2x) \to 3(1+2\cdot 0) = 3(1+0) = 3\cdot 1 = 3$

The denominator approaches 0, because $\sin(0)=0$

Moreover, we have

$\displaystyle \lim_{x\to 0^-} \sin(x) = 0^-,\quad \displaystyle \lim_{x\to 0^+} \sin(x) = 0^+$

So, the limit does not exist, because left and right limits are different:

$\displaystyle \lim_{x\to 0^-} \dfrac{3(1+2x)}{\sin(x)}= \dfrac{3}{0^-} = -\infty,\quad \displaystyle \lim_{x\to 0^+}\dfrac{3(1+2x)}{\sin(x)}= \dfrac{3}{0^+} = +\infty$