What is the limit as x approaches-3 of (1-6x)/(3+x) ?

The limit as x approaches-3 of (1-6x)/(3+x) is diverges

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limit as x approaches-3 of (1-6x)/(3+x)

x \to - 3 lim (\frac{1 - 6 x}{3 + x})

diverges

Solution steps

x \to - 3 lim (\frac{1 - 6 x}{3 + x})

x \to a - lim f (x) \neq = x \to a + lim f (x)

then the limit does not exist

x \to - 3 + lim (\frac{1 - 6 x}{3 + x}) = \infty

x \to - 3 - lim (\frac{1 - 6 x}{3 + x}) = - \infty

= diverges

\int_{0}^{1} \frac{x ^{2}}{x ^{3} + 1} d x

\frac{d}{d x} (\frac{sin ( \frac{1}{x} ) x ^{3}}{ln ( x )})

a re a 1 - x^{2}, x^{6} - x + 1

x \to - \infty lim (x^{4} + x)

sin (2 x) d x + cos (8 y) d y = 0

What is the limit as x approaches-3 of (1-6x)/(3+x) ?
The limit as x approaches-3 of (1-6x)/(3+x) is diverges