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

3SAT
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user315918
  • 599
  • No you can use whenever you have the forms $\frac{0}{0}, \frac{\infty}{\infty}$. However this limit is of the form $\frac{7}{0}$ which does not exist as left-approach is not equal to the right-approach. – crbah Mar 08 '16 at 15:27
  • if the numerator was also zero, then you could use l'Hopital rule. Here the function diverges in the vicinity of 1. – Math-fun Mar 08 '16 at 15:27
  • @Math-fun so it's $\infty$ – user315918 Mar 08 '16 at 15:31
  • for $x \to 1^+$ you get $-\infty$ and for $\infty$ for the left limit. – Math-fun Mar 08 '16 at 15:41
  • @user315918 don't forget to accept one of the answers clicking the 'V' in the left side of the answer, and also to vote – 3SAT Mar 08 '16 at 19:26

2 Answers2

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

Whiz_Geek
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0

The limit does not exist:

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

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

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

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

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

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

3SAT
  • 7,512
  • $\lim_{x \to a} \frac{f(x)}{g(x)} \ne \lim_{x \to a} f(x) \lim_{x \to a} \frac{1}{g(x)}$ if $\lim_{x \to a} g(x) = 0$ – crbah Mar 08 '16 at 15:37
  • Basic Limit Laws - Division Law : http://math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/lHopital/limit_laws.html – crbah Mar 08 '16 at 15:42
  • In fact the same constraint is still valid for the product rule. $\lim_{x \to a} \frac{1}{g(x)}$ should be finite. Try to use the rule for $\lim_{x \to 0} x.\frac{2}{x}$ – crbah Mar 08 '16 at 15:48