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Question Number 134822 by abdurehime last updated on 07/Mar/21

Answered by greg_ed last updated on 07/Mar/21

$$\mathrm{p}=\underset{x\to 1}{\lim }\frac{x^4-\sqrt{x}}{\sqrt{x}-1}$$$$\mathbf{by}{\mathbf{L}}^{\prime }{\mathbf{Hopital}}^{\prime }\mathbf{s}\mathbf{rule}$$$$\mathrm{p}=\underset{x\to 1}{\lim }\frac{4x^3-\frac{1}{2\sqrt{x}}}{\frac{1}{2\sqrt{x}}}$$$$\mathbf{plug}\mathbf{in}\mathit{x}=1$$$$\underset{x\to 1}{\lim }\frac{x^4-\sqrt{x}}{\sqrt{x}-1}=7$$

Answered by EDWIN88 last updated on 07/Mar/21

$$\underset{x\to 1}{\lim }\frac{{\mathrm{x}}^4-\sqrt{\mathrm{x}}}{\sqrt{\mathrm{x}}-1}$$$$\mathrm{let}\sqrt{\mathrm{x}}=\mathrm{u}\mathrm{then}\underset{\mathrm{u}\to 1}{\lim }\frac{{\mathrm{u}}^8-\mathrm{u}}{\mathrm{u}-1}=\underset{\mathrm{u}\to 1}{\lim }\frac{\mathrm{u}({\mathrm{u}}^7-1)}{\mathrm{u}-1}$$$$=\underset{\mathrm{u}\to 1}{\lim }\mathrm{u}({\mathrm{u}}^6+{\mathrm{u}}^5+{\mathrm{u}}^4+{\mathrm{u}}^3+{\mathrm{u}}^2+\mathrm{u}+1)$$$$=1\times \underset{7\mathrm{times}}{\underbrace{(1+1+1+...+1)}}=7\times 1=7$$

Answered by EDWIN88 last updated on 07/Mar/21

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{\sqrt{{9\mathrm{x}}^4+1}}{{\mathrm{x}}^2-3\mathrm{x}+5}=\underset{x\to \mathrm{\infty }}{\lim }\frac{{\mathrm{x}}^2\sqrt{9+\frac{1}{{\mathrm{x}}^4}}}{{\mathrm{x}}^2(1-\frac{3}{\mathrm{x}}+\frac{5}{{\mathrm{x}}^2})}$$$$=\underset{\mathrm{x}\to \mathrm{\infty }}{\lim }\frac{\sqrt{9+\frac{1}{{\mathrm{x}}^2}}}{1-\frac{3}{\mathrm{x}}+\frac{5}{{\mathrm{x}}^2}};\mathrm{let}\frac{1}{\mathrm{x}}=\mathrm{t}\mathrm{and}\mathrm{t}\to 0$$$$=\underset{\mathrm{t}\to 0}{\lim }\frac{\sqrt{9+{\mathrm{t}}^2}}{1-3\mathrm{t}+{5\mathrm{t}}^2}=\frac{3}{1}=3$$

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