I have $\lim\limits_{x \to 2 \pm 0} \frac{(x+1)^2}{2-x}$
I calculated $\lim\limits_{x \to 2 + 0} \frac{(x+1)^2}{2-x}=\lim\limits_{x \to 2 - 0} \frac{(x+1)^2}{2-x}=\lim\limits_{x \to 2 } \frac{(x+1)^2}{2-x}=\frac90=\infty$
Or is it wrong?
Calculate limit $\lim_{x \to 2 \pm 0} \frac{(x+1)^2}{2-x}$ without L'Hopital's rule
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limits-without-lhopital
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0yeah seems to be fine! – 2017-01-10
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3isn't $+-\infty$? – 2017-01-10
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0The limit still $+-\infty$ you are mulplying it by $\frac{x^2}{x^2}=1$ you are just multiplyng by $1$ – 2017-01-10
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0R.MCM thanks... – 2017-01-10
2 Answers
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No, this isn't correct. The left- and right-side limits are not equal.
$\lim \limits_{x\to 2^-}\frac{(x+1)^2}{2-x} = \left[\frac{9}{0^+}\right] = +\infty$
$\lim \limits_{x\to 2^+}\frac{(x+1)^2}{2-x} = \left[\frac{9}{0^-}\right] = -\infty$
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For a different look at it, set $x-2=t$ to get $$ - \lim_{t \to 0} \frac{(t+3)^2}{t} = -6 - \lim_{t \to 0} \frac{9}{t} $$ Clearly as $t \to 0$, this limit does not exist.