What is the limit as x approaches 1+of 1/(x^2-4x+3) ?

The limit as x approaches 1+of 1/(x^2-4x+3) is -infinity

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

limit as x approaches 1+of 1/(x^2-4x+3)

x \to 1 + lim (\frac{1}{x ^{2} - 4 x + 3})

- \infty

Solution steps

x \to 1 + lim (\frac{1}{x ^{2} - 4 x + 3})

For

x

approaching

1

from the right

, x > 1 \Rightarrow x^{2} - 4 x + 3 < 0

The denominator is a negative quantity approaching 0 from the left

= - \infty

\frac{\partial}{\partial x} (tan (x^{3} + y))

\int 5 arcsin (x) d x

y^{^{'''}} + 3 y^{^{''}} + 2 y^{^{'}} = 0

x \to 0 lim (\frac{8 ^{x} - 1}{x})

\frac{\partial}{\partial y} (\frac{x ^{4}}{6 y} + \frac{12 y ^{4}}{x})

What is the limit as x approaches 1+of 1/(x^2-4x+3) ?
The limit as x approaches 1+of 1/(x^2-4x+3) is -infinity