4

How to solve this limit $$\lim_{x\rightarrow 0}{\frac{(2+x)^x-2^x}{x^2}}$$ using the concept of equivalent functions?

For example, if $x\rightarrow 0 $ function $\sin x$ is equivalent to $x$, $\ln(1+x)\sim x$, $a^x-1 \sim x \ln a$, etc.

Alexander Side
  • 490

2 Answers2

3

What OP calls "equivalent functions" are normally called Taylor series, expanded in this case at $0$.

$$2^x=1+x\log 2 + \frac{1}{2}x^2\log^22+O(x^3)$$ $$(2+x)^x=1+x\log 2 + \frac{1}{2}x^2(1+\log^22)+O(x^3)$$

Hence the fraction becomes $$\frac{x^2/2+O(x^3)}{x^2}=\frac{1}{2}+O(x)$$ Taking limits as $x\to 0$ gives an answer of $\frac{1}{2}$. To find the Taylor expansions quickly you can use alpha.

vadim123
  • 82,796
3

we have $$\lim_{x\to 0}{\frac{(2+x)^x-2^x}{x^2}}=\lim_{x\to 0}2^x\lim_{x\to 0}{\frac{(1+\frac{x}{2})^x-1}{x^2}}=\lim_{x\to 0}{\frac{e^{x\ln(1+\frac{x}{2})}-1}{x^2}}$$ therefore we have that$$\lim_{x\to 0}{\frac{(2+x)^x-2^x}{x^2}}\sim\lim_{x\to 0}{\frac{{x\ln(1+\frac{x}{2})}}{x^2}}=\frac{1}{2}$$

user62498
  • 3,556