Evaluate $\lim_{x \to 1} \frac {2x+5}{x^{2}-4x+3}$

Question

$$\lim_{x \to 1} \frac {2x+5}{x^{2}-4x+3}$$ We have $0$ in denominator i don't know how to use l'Hoptial rule and i think we can't use the rule of polynomial functions

Answer 1

The limit does not exist:

$$\lim_{x \to 1^{+}} \frac {2x+5}{x^{2}-4x+3}$$

$$\lim_{x \to 1^{+}} (2x+5) \lim_{x \to 1^{+}}\frac {1}{x^{2}-4x+3}$$

$$=7\cdot\lim_{x \to 1^{+}}\frac {1}{\underbrace{x^{2}-4x+3}_{\to 0^-}}=-\infty$$

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$$\lim_{x \to 1^{-}} \frac {2x+5}{x^{2}-4x+3}$$

$$\lim_{x \to 1^{-}} (2x+5) \lim_{x \to 1^{+}}\frac {1}{x^{2}-4x+3}$$

$$=7\cdot\lim_{x \to 1^{-}}\frac {1}{\underbrace{x^{2}-4x+3}_{\to 0^+}}=+\infty$$

Answer 2

This limit does not exist. left hand limit is - infinity. Right hand limit is +infinity