Question

# Find the limits. Write infty or -infty where appropriate. lim_{xrightarrow0^-}frac{x^2-3x+2}{x^3-4x}

Limits and continuity
Find the limits. Write $$\infty$$ or $$-\infty$$ where appropriate. $$\lim_{x\rightarrow0^-}\frac{x^2-3x+2}{x^3-4x}$$

2021-02-01

First let's write the limit
$$\lim_{x\rightarrow0^-}\frac{x^2-3x+2}{x^3-4x}$$
$$=\lim_{h\rightarrow0}\frac{x^2-3x+2}{x^3-4x}$$
$$=\lim_{h\rightarrow0}\frac{-h^2-3(-h)+2}{-h^3+4h}$$
$$=\lim_{h\rightarrow0}\frac{h^2+3g+2}{-h^3+4h}$$
Then, since h tends to 0, it must be fractional and less than 1
As $$h < 1$$
$$h^3$$
Hence, $$h^3<4h$$
$$4h-h^3>0$$
Hence, the denominator in the limit above approaches $$0^+$$
$$Limit=\frac{0+0+2}{0^+}=\infty$$
Result:$$\infty$$