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Question Number 131079 by mathlove last updated on 01/Feb/21

$$\underset{h\to 0}{\lim }\frac{e^{2\sqrt{2}x}{e}^{2\sqrt{2}h}-e^{2\sqrt{2}x}}{h}=?$$

Commented by EDWIN88 last updated on 01/Feb/21

$$\underset{h\to 0}{\lim }\frac{e^{2\sqrt{2}(x+h)}-e^{2\sqrt{2}x}}{h}=f{}^{\prime }\left(x\right)$$$$wheref\left(x\right)=e^{2\sqrt{2}x}$$$$lety=e^{2\sqrt{2}x}\Rightarrow \ln \left(y\right)=2\sqrt{2}x$$$$\frac{y^{\prime }}{y}=2\sqrt{2}\Rightarrow y^{\prime }=2\sqrt{2}{e}^{2\sqrt{2}x}$$$$so\underset{h\to 0}{\lim }\frac{e^{2\sqrt{2}(x+h)}-e^{2\sqrt{2}x}}{h}=2\sqrt{2}{e}^{2\sqrt{2}x}$$$$$$

Answered by EDWIN88 last updated on 01/Feb/21

$$e^{2\sqrt{2}x}\times \underset{h\to 0}{\lim }\frac{e^{2\sqrt{2}h}-1}{h}$$

Commented by mathlove last updated on 01/Feb/21

$$and$$$$\underset{h\to 0}{\lim }\frac{e^{2\sqrt{2}h}-1}{h}=?$$

Commented by Dwaipayan Shikari last updated on 01/Feb/21

$$\underset{x\to 0}{\lim }\frac{e^{2\sqrt{2}x}-1}{x}=\frac{1+2\sqrt{2}x-1}{x}=2\sqrt{2}\underset{x\to 0}{\lim }{e}^x=1+x$$

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