Fraction Limit

Calculus Level 2

lim x 0 x x = ? \large \lim _{ x\rightarrow 0} \frac {|x|}x = ?

0 \infty -\infty 1 -1 Does not exist

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1 solution

Chew-Seong Cheong
Oct 29, 2017

Relevant wiki: L'Hopital's Rule - Basic

L + = lim x 0 + x x = lim x 0 + x x A 0/0 case, L’H o ˆ pital’s rule applies. = lim x 0 + 1 1 Differentiate up and down w.r.t. x = 1 \begin{aligned} L_+ & = \lim_{x \to 0^+} \frac {|x|}x \\ & = \lim_{x \to 0^+} \frac xx & \small \color{#3D99F6} \text{A 0/0 case, L'Hôpital's rule applies.} \\ & = \lim_{x \to 0^+} \frac 11 & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }x \\ & = 1 \end{aligned}

L = lim x 0 x x = lim x 0 x x A 0/0 case, L’H o ˆ pital’s rule applies. = lim x 0 1 1 Differentiate up and down w.r.t. x = 1 \begin{aligned} L_- & = \lim_{x \to 0^-} \frac {|x|}x \\ & = \lim_{x \to 0^-} \frac {-x}x & \small \color{#3D99F6} \text{A 0/0 case, L'Hôpital's rule applies.} \\ & = \lim_{x \to 0^-} \frac {-1}1 & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }x \\ & = -1 \end{aligned}

Since L + L L_+ \ne L_- , the limit does not exist .

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