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Question Number 184794 by Spillover last updated on 11/Jan/23

$$Evaluate$$$$\underset{x\to 2}{\lim }\frac{x^5-32}{x^3-8}$$

Commented by MJS_new last updated on 11/Jan/23

$$\frac{20}{3}$$

Commented by Spillover last updated on 11/Jan/23

$$thankyou$$

Answered by Spillover last updated on 12/Jan/23

$$\underset{x\to 2}{\lim }\frac{x^5-2^5}{x^3-2^3}=\underset{x\to 2}{\lim }\frac{\frac{x^5-2^5}{x-2}}{\frac{x^3-2^3}{x-2}}$$$$from$$$$\underset{x\to a}{\lim }\frac{x^n-a^n}{x-a}={nax}^{n-1}$$$$\mathrm{li}\underset{x\to 2}{\mathrm{m}}\frac{\frac{x^5-2^5}{x-2}}{\frac{x^3-2^3}{x-2}}=\underset{x\to 2}{\lim }\frac{x^5-2^5}{x-2}\mathbf{÷}\underset{x\to 2}{\lim }\frac{x^3-2^3}{x-2}$$$$\left(5\times 2^4\right)\mathbf{÷}\left(3\times 2^2\right)=\frac{20}{3}$$$$$$

Answered by Rasheed.Sindhi last updated on 12/Jan/23

$$\underset{x\to 2}{\lim }\frac{x^5-32}{x^3-8}$$$$=\underset{x\to 2}{\lim }\frac{\cancel{(x-2)}(x^4+2x^3+4x^2+8x+16)}{\cancel{(x-2)}(x^2+2x+4)}$$$$=\frac{\underset{x\to 2}{\lim }(x^4+2x^3+4x^2+8x+16)}{\underset{x\to 2}{\lim }(x^2+2x+4)}$$$$=\frac{2^4+2\cdot 2^3+4\cdot 2^2+8\cdot 2+16}{2^2+2\cdot 2+4}$$$$=\frac{5\cdot 2^4}{3\cdot 2^2}=\frac{5}{3}\cdot 2^{4-2}=\frac{20}{3}$$

Commented by Spillover last updated on 12/Jan/23

$$niceapproach$$

Answered by BaliramKumar last updated on 14/Jan/23

$$\underset{x\to 2}{\lim }\frac{x^5-32}{x^3-8}$$$$\underset{x\to 2}{\lim }\frac{5x^4}{3x^2}=\frac{5x^2}{3}=\frac{5{\left(2\right)}^2}{3}=\frac{20}{3}$$$$$$

Answered by ajfour last updated on 14/Jan/23

$$\underset{x\to a}{\lim }\frac{x^n-a^n}{x-a}={na}^{n-1}$$$$hence=\frac{5\left(2^4\right)}{3\left(2^2\right)}=\frac{20}{3}$$

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