$$ x_{0}=1 $$ רטว $$ f(x)=\left\{\begin{array}{ll}\frac{|x-2|}{x-2} & x

Question

$$ x_{0}=1 $$ רטว $$ f(x)=\left\{\begin{array}{ll}\frac{|x-2|}{x-2} & x<1 \ \frac{e^{2 x-2}-1}{x-1} & x \geq 1\end{array}ight. $$

Reid Ayala · Accepted Answer

The limit of $$ f(x) $$ as $$ x $$ approaches 1 does not exist because the left-hand limit is $$-1$$ and the right-hand limit is $$2$$, which are not equal.

\( x_{0}=1 \) רטว \( f(x)=\left\{\begin{array}{ll}\frac{|x-2|}{x-2} & x<1 \\ \frac{e^{2 x-2}-1}{x-1} & x \geq 1\end{array}\right. \)

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Solución paso a paso

\( x_{0}=1 \) רטว \( f(x)=\left\{\begin{array}{ll}\frac{|x-2|}{x-2} & x<1 \\ \frac{e^{2 x-2}-1}{x-1} & x \geq 1\end{array}\right. \)

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Respuesta rápida

Solución paso a paso

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