Question Number 133474 by Abdoulaye last updated on 22/Feb/21
Commented by Abdoulaye last updated on 22/Feb/21
$${please} \\ $$
Answered by Olaf last updated on 22/Feb/21
![e^(√(x+1)) −e^(√x) = e^(((√(x+1))+(√x))/2) [e^(((√(x+1))−(√x))/2) −e^(−(((√(x+1))−(√x))/2)) ] = 2e^(((√(x+1))+(√x))/2) sinh((((√(x+1))−(√x))/2)) = 2e^(((√(x+1))+(√x))/2) sinh((1/(2((√(x+1))+(√x))))) ∼_∞ 2e^(√x) sinh((1/(4(√x)))) ∼_∞ 2e^(√x) ×(1/(4(√x))) = ∞](Q133502.png)
$${e}^{\sqrt{{x}+\mathrm{1}}} −{e}^{\sqrt{{x}}} \:=\:{e}^{\frac{\sqrt{{x}+\mathrm{1}}+\sqrt{{x}}}{\mathrm{2}}} \left[{e}^{\frac{\sqrt{{x}+\mathrm{1}}−\sqrt{{x}}}{\mathrm{2}}} −{e}^{−\frac{\sqrt{{x}+\mathrm{1}}−\sqrt{{x}}}{\mathrm{2}}} \right] \\ $$$$=\:\mathrm{2}{e}^{\frac{\sqrt{{x}+\mathrm{1}}+\sqrt{{x}}}{\mathrm{2}}} \mathrm{sinh}\left(\frac{\sqrt{{x}+\mathrm{1}}−\sqrt{{x}}}{\mathrm{2}}\right) \\ $$$$=\:\mathrm{2}{e}^{\frac{\sqrt{{x}+\mathrm{1}}+\sqrt{{x}}}{\mathrm{2}}} \mathrm{sinh}\left(\frac{\mathrm{1}}{\mathrm{2}\left(\sqrt{{x}+\mathrm{1}}+\sqrt{{x}}\right)}\right) \\ $$$$\underset{\infty} {\sim}\:\mathrm{2}{e}^{\sqrt{{x}}} \mathrm{sinh}\left(\frac{\mathrm{1}}{\mathrm{4}\sqrt{{x}}}\right)\:\underset{\infty} {\sim}\:\mathrm{2}{e}^{\sqrt{{x}}} ×\frac{\mathrm{1}}{\mathrm{4}\sqrt{{x}}}\:=\:\infty \\ $$