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Question Number 170221 by mathlove last updated on 18/May/22

$$\underset{x\to 0^{+}}{\lim }\left(\frac{{\left(\sqrt{x}\right)}^{\sqrt{x}}-x^x}{{\left(\sqrt{x}\right)}^x-x^{\sqrt{x}}}\right)=?$$$$helpme$$

Commented by mathlove last updated on 19/May/22

$$anyoneisanswer$$

Answered by qaz last updated on 19/May/22

$$\underset{\mathrm{x}\to 0^{+}}{\lim }\frac{{\left(\sqrt{\mathrm{x}}\right)}^{\sqrt{\mathrm{x}}}-{\mathrm{x}}^{\mathrm{x}}}{{\left(\sqrt{\mathrm{x}}\right)}^{\mathrm{x}}-{\mathrm{x}}^{\sqrt{\mathrm{x}}}}=\underset{\mathrm{x}\to 0^{+}}{\lim }\frac{{\mathrm{e}}^{\sqrt{\mathrm{x}}\ln \sqrt{\mathrm{x}}}-{\mathrm{e}}^{\mathrm{xlnx}}}{{\mathrm{e}}^{\mathrm{xln}\sqrt{\mathrm{x}}}-{\mathrm{e}}^{\sqrt{\mathrm{x}}\mathrm{lnx}}}=\underset{\mathrm{x}\to 0^{+}}{\lim }\frac{{\mathrm{e}}^{\mathrm{xlnx}}({\mathrm{e}}^{\sqrt{\mathrm{x}}\ln \sqrt{\mathrm{x}}-\mathrm{xlnx}}-1)}{{\mathrm{e}}^{\sqrt{\mathrm{x}}\mathrm{lnx}}({\mathrm{e}}^{\mathrm{xln}\sqrt{\mathrm{x}}-\sqrt{\mathrm{x}}\mathrm{lnx}}-1)}$$$$=\underset{\mathrm{x}\to 0^{+}}{\lim }\frac{{\mathrm{e}}^{(\mathrm{x}-\sqrt{\mathrm{x}})\mathrm{lnx}}(\sqrt{\mathrm{x}}\ln \sqrt{\mathrm{x}}-\mathrm{xlnx})}{\mathrm{xln}\sqrt{\mathrm{x}}-\sqrt{\mathrm{x}}\mathrm{lnx}}=\underset{\mathrm{x}\to 0^{+}}{\lim }\frac{\ln \sqrt{\mathrm{x}}-\sqrt{\mathrm{x}}\mathrm{lnx}}{\sqrt{\mathrm{x}}\ln \sqrt{\mathrm{x}}-\mathrm{lnx}}$$$$=\underset{\mathrm{x}\to 0^{+}}{\lim }\frac{\frac{1}{2}\mathrm{lnx}-\sqrt{\mathrm{x}}\mathrm{lnx}}{\frac{1}{2}\sqrt{\mathrm{x}}\mathrm{lnx}-\mathrm{lnx}}=\underset{\mathrm{x}\to 0^{+}}{\lim }\frac{\frac{1}{2}-\sqrt{\mathrm{x}}}{\frac{1}{2}\sqrt{\mathrm{x}}-1}=-\frac{1}{2}$$

Commented by mathlove last updated on 19/May/22

$$\mathcal{NICE}$$

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