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LIMITS...PLEASE HELP
- Thread starter nikchic5
- Start date
Hello, nikchic5!
\(\displaystyle \L\;\;\lim_{z\to0}(1\,+\,z)^{\frac{1}{z}}\;=\;e\)
\(\displaystyle \text{We have: }\L\,(1\,+\,2x)^{\frac{1}{x}}\)
\(\displaystyle \text{Multply the exponent by }\L\frac{2}{2}:\;\;(1\,+\,2x)^{\frac{1}{x}\cdot\frac{2}{2}} \;=\;(1\,+\,2x)^{\frac{1}{2x}\cdot2} \;=\;\left[(1\,+\,2x)^{\frac{1}{2x}}\right]^2\)
\(\displaystyle \text{Therefore: }\L\:\lim_{2x\to0^+}\left[(1\,+\,2x)^{\frac{1}{2x}}\right]^2 \;= \;\left[\lim_{2x\to0^+} (1\,+\,2x)^{\frac{1}{2x}}\right]^2 \;= \;\L e^2\)
Are you familiar with this defintion of \(\displaystyle e\) ?\(\displaystyle \L\lim_{x\to0^+}(1\,+\,2x)^{\frac{1}{x}}\)
\(\displaystyle \L\;\;\lim_{z\to0}(1\,+\,z)^{\frac{1}{z}}\;=\;e\)
\(\displaystyle \text{We have: }\L\,(1\,+\,2x)^{\frac{1}{x}}\)
\(\displaystyle \text{Multply the exponent by }\L\frac{2}{2}:\;\;(1\,+\,2x)^{\frac{1}{x}\cdot\frac{2}{2}} \;=\;(1\,+\,2x)^{\frac{1}{2x}\cdot2} \;=\;\left[(1\,+\,2x)^{\frac{1}{2x}}\right]^2\)
\(\displaystyle \text{Therefore: }\L\:\lim_{2x\to0^+}\left[(1\,+\,2x)^{\frac{1}{2x}}\right]^2 \;= \;\left[\lim_{2x\to0^+} (1\,+\,2x)^{\frac{1}{2x}}\right]^2 \;= \;\L e^2\)
Soroban's method is very nice and, no doubt, the best way, but here's another way.....if I may.
\(\displaystyle \L\\\lim_{x\to\0^{+}}(1+2x)^{\frac{1}{x}}\)
\(\displaystyle \L\\\lim_{x\to\0^{+}}e^{\frac{1}{x}ln(1+2x)}\)
\(\displaystyle \L\\e^{(\lim_{x\to\0^{+}}(\frac{ln(1+2x)}{x}))}\)
Using L'Hopitals's Rule and differentiating the numerator and denominator,
we get:
\(\displaystyle \L\\e^{(\lim_{x\to\0^{+}}(\frac{2}{2x+1}))}\)
\(\displaystyle \L\\e^{(2\lim_{x\to\0^{+}}(\frac{1}{1+2x}))}\)
As you can see, the limit inside the parentheses goes to 1, and we have:
\(\displaystyle \H\\e^{2}\)
\(\displaystyle \L\\\lim_{x\to\0^{+}}(1+2x)^{\frac{1}{x}}\)
\(\displaystyle \L\\\lim_{x\to\0^{+}}e^{\frac{1}{x}ln(1+2x)}\)
\(\displaystyle \L\\e^{(\lim_{x\to\0^{+}}(\frac{ln(1+2x)}{x}))}\)
Using L'Hopitals's Rule and differentiating the numerator and denominator,
we get:
\(\displaystyle \L\\e^{(\lim_{x\to\0^{+}}(\frac{2}{2x+1}))}\)
\(\displaystyle \L\\e^{(2\lim_{x\to\0^{+}}(\frac{1}{1+2x}))}\)
As you can see, the limit inside the parentheses goes to 1, and we have:
\(\displaystyle \H\\e^{2}\)