Question Number 215889 by hardmath last updated on 20/Jan/25
$$\mathrm{Find}:\:\:\:\underset{\boldsymbol{\mathrm{h}}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{x}\:+\:\mathrm{h}\right)^{\mathrm{3}} \:+\:\mathrm{x}^{\mathrm{3}} }{\mathrm{h}}\:=\:? \\ $$
Commented by mr W last updated on 20/Jan/25
$$\rightarrow\frac{\mathrm{2}{x}^{\mathrm{3}} }{\mathrm{0}}\rightarrow\infty \\ $$
Commented by hardmath last updated on 20/Jan/25
$$ \\ $$Dear professor, 2x why, please write clearly
Commented by mr W last updated on 20/Jan/25
$${when}\:{h}\rightarrow\mathrm{0},\:\left({x}+{h}\right)^{\mathrm{3}} +{x}^{\mathrm{3}} \rightarrow{x}^{\mathrm{3}} +{x}^{\mathrm{3}} =\mathrm{2}{x}^{\mathrm{3}} \\ $$
Commented by mr W last updated on 20/Jan/25
$${question}\:{makes}\:{no}\:{much}\:{sense}. \\ $$
Commented by hardmath last updated on 20/Jan/25
$$\mathrm{thankyou}\:\mathrm{dear}\:\mathrm{professor} \\ $$
Answered by MathematicalUser2357 last updated on 21/Jan/25
$$\: \\ $$
Commented by mr W last updated on 21/Jan/25
$$\frac{\left({x}+{h}\right)^{\mathrm{3}} +{x}^{\mathrm{3}} }{{h}}\neq\frac{\left({x}+{h}\right)^{\mathrm{3}} −{x}^{\mathrm{3}} }{{h}} \\ $$
Commented by MathematicalUser2357 last updated on 21/Jan/25
$${i}'{m}\:{sorry}\:{for}\:\frac{\left({x}+{h}\right)^{\mathrm{3}} +{x}^{\mathrm{3}} }{{h}}=\frac{\left({x}+{h}\right)^{\mathrm{3}} −{x}^{\mathrm{3}} }{{h}}. \\ $$$${so}\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({x}+{h}\right)^{\mathrm{3}} +{x}^{\mathrm{3}} }{{h}}=\frac{\mathrm{2}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} {h}+\mathrm{3}{xh}^{\mathrm{2}} +{h}^{\mathrm{3}} }{{h}}=\frac{\mathrm{2}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} ×\mathrm{0}+\mathrm{3}{x}×\mathrm{0}^{\mathrm{2}} +\mathrm{0}^{\mathrm{3}} }{\mathrm{0}}=\frac{\mathrm{2}{x}^{\mathrm{3}} }{\mathrm{0}}={Indeterminate}! \\ $$
Commented by mr W last updated on 21/Jan/25
$${yes}. \\ $$
Commented by hardmath last updated on 21/Jan/25
$$\mathrm{thankyou}\:\mathrm{dear}\:\mathrm{professors} \\ $$