Final answer to the problem
Step-by-step Solution
How should I solve this problem?
- Scegliere un'opzione
- Prodotto di binomi con termine comune
- Metodo FOIL
- Sostituzione di Weierstrass
- Dimostrare dal LHS (lato sinistro)
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Apply the formula: $derivdef\left(x\right)$$=\lim_{h\to0}\left(\frac{eval\left(x,var+h\right)-x}{h}\right)$, where $derivdefx=derivdef\left(\ln\left(x\right)\right)$ and $x=\ln\left(x\right)$
Apply the formula: $\ln\left(a\right)-\ln\left(b\right)$$=\ln\left(\frac{a}{b}\right)$, where $a=x+h$ and $b=x$
Apply the formula: $\frac{a}{b}$$=\frac{1}{b}a$, where $a=\ln\left(\frac{x+h}{x}\right)$ and $b=h$
Apply the formula: $a\ln\left(x\right)$$=\ln\left(x^a\right)$, where $a=\frac{1}{h}$ and $x=\frac{x+h}{x}$
Expand the fraction $\left(\frac{x+h}{x}\right)$ into $2$ simpler fractions with common denominator $x$
Simplify the resulting fractions
Apply the formula: $\lim_{h\to0}\left(\ln\left(\left(1+\frac{h}{x}\right)^{\frac{1}{h}}\right)\right)$$=\lim_{n\to\infty }\left(\ln\left(\left(1+\frac{1}{n}\right)^{\frac{n}{x}}\right)\right)$, where $h/x=\frac{h}{x}$, $1+h/x=1+\frac{h}{x}$, $h->0=h\to0$ and $1/h=\frac{1}{h}$
Apply the formula: $a^{\frac{b}{c}}$$=\left(a^b\right)^{\frac{1}{c}}$, where $a=1+\frac{1}{n}$, $b=n$, $c=x$ and $b/c=\frac{n}{x}$
Apply the formula: $\ln\left(x^a\right)$$=a\ln\left(x\right)$, where $a=\frac{1}{x}$ and $x=\left(1+\frac{1}{n}\right)^n$
Apply the formula: $\lim_{x\to c}\left(ab\right)$$=a\lim_{x\to c}\left(b\right)$, where $a=\frac{1}{x}$, $b=\ln\left(\left(1+\frac{1}{n}\right)^n\right)$, $c=\infty $ and $x=n$
Apply the formula: $\lim_{x\to c}\left(\ln\left(a\right)\right)$$=\ln\left(\lim_{x\to c}\left(a\right)\right)$, where $a=\left(1+\frac{1}{n}\right)^n$, $c=\infty $ and $x=n$
Apply the formula: $\lim_{x\to\infty }\left(\left(1+\frac{a}{x}\right)^x\right)$$=e^a$, where $a=1$ and $x=n$
Apply the formula: $\ln\left(x\right)$$=logf\left(x,e\right)$, where $x=e^1$
