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

Question

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

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Genesis 1 tuần 2021-09-02T13:07:02+00:00 2 Answers 3 views 0

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    0
    2021-09-02T13:08:13+00:00

    Đáp án:

    \(\lim\limits_{x\to 3^+}\dfrac{|2-x|}{2x^2 – 5x+2}=\dfrac15\) 

    Giải thích các bước giải:

    \(\begin{array}{l}
    \quad \lim\limits_{x\to 3^+}\dfrac{|2-x|}{2x^2 – 5x+2}\\
    = \lim\limits_{x\to 3^+}\dfrac{x-2}{(x-2)(2x-1)}\\
    = \lim\limits_{x\to 3^+}\dfrac{1}{2x-1}\\
    = \dfrac{1}{2.3 -1}\\
    = \dfrac{1}{5}
    \end{array}\)

    0
    2021-09-02T13:08:46+00:00

    Đáp án:

     

    Giải thích các bước giải:

    Nếu  `lim_{x->3^+} \frac{|2-x|}{2x² -5+2}`

    `= lim_{x->3^+} \frac{x-2}{2x²-5+2}`

    `= \frac{3-2}{2.3²-5+2}`

    ` =\frac{1}{15}`

    Nếu `lim_{x->3^+} \frac{|2-x|}{2x²-5x+2}`

    `= lim_{x->3^+} \frac{x-2}{2x²-5x+2}`

    `= \frac{3-2}{2.3²-5.3+2}`

    `=1/5`

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