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Question Number 215889 by hardmath last updated on 20/Jan/25

$$\mathrm{Find}:\underset{\mathbf{h}\to 0}{\lim }\frac{{(\mathrm{x}+\mathrm{h})}^3+{\mathrm{x}}^3}{\mathrm{h}}=?$$

Commented by mr W last updated on 20/Jan/25

$$\to \frac{2x^3}{0}\to \mathrm{\infty }$$

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

$$whenh\to 0,{(x+h)}^3+x^3\to x^3+x^3=2x^3$$

Commented by mr W last updated on 20/Jan/25

$$questionmakesnomuchsense.$$

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

$$\frac{{(x+h)}^3+x^3}{h}\ne \frac{{(x+h)}^3-x^3}{h}$$

Commented by MathematicalUser2357 last updated on 21/Jan/25

$$i^{\prime }msorryfor\frac{{(x+h)}^3+x^3}{h}=\frac{{(x+h)}^3-x^3}{h}.$$$$so\underset{h\to 0}{\lim }\frac{{(x+h)}^3+x^3}{h}=\frac{2x^3+3x^{2}h+3{xh}^2+h^3}{h}=\frac{2x^3+3x^2\times 0+3x\times 0^2+0^3}{0}=\frac{2x^3}{0}=Indeterminate!$$

Commented by mr W last updated on 21/Jan/25

$$yes.$$

Commented by hardmath last updated on 21/Jan/25

$$\mathrm{thankyou}\mathrm{dear}\mathrm{professors}$$

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