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Question Number 184790 by Spillover last updated on 11/Jan/23

$$Evaluate$$$$\underset{x\to 0}{\lim }\frac{{(1+x)}^6-1}{{(1+x)}^5-1}$$

Commented by MJS_new last updated on 11/Jan/23

$${\mathrm{l}}^{\prime }\mathrm{H}\widehat{\mathrm{o}pital}\Rightarrow \frac{6}{5}$$

Commented by Spillover last updated on 11/Jan/23

$$yourright$$

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

$$Let1+x=y,$$$$x\to 0\Rightarrow y\to 1$$$$\underset{y\to 1}{\lim }\frac{y^6-1}{y^5-1}=\underset{y\to 1}{\lim }\frac{\cancel{(y-1)}(y+1)(y^2-y+1)(y^2+y+1)}{\cancel{(y-1)}(y^4+y^3+y^2+y+1)}$$$$=\underset{y\to 1}{\lim }\frac{(y+1)(y^2-y+1)(y^2+y+1)}{(y^4+y^3+y^2+y+1)}$$$$=\frac{(1+1)(1^2-1+1)(1^2+1+1)}{1^4+1^3+1^2+1+1}=\frac{6}{5}$$

Answered by Spillover last updated on 12/Jan/23

$$1+x=yx\to 0y=1$$$$\underset{y\to 1}{\lim }\frac{y^6-1}{y^5-1}=\underset{y\to 1}{\lim }\frac{\frac{y^6-1}{y-1}}{\frac{y^5-1}{y-1}}$$$$from\underset{x\to 0}{\lim }\frac{x^n-a^n}{x-a}={na}^{n-1}$$$$\mathrm{li}\underset{y\to 1}{\mathrm{m}}\frac{\frac{y^6-1}{y-1}}{\frac{y^5-1}{y-1}}=\frac{6\times 1^{6-1}}{5\times 1^{5-1}}=\frac{6}{5}$$

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