# Evaluate the following limit. lim_{xrightarrow0^+}(1+5x)^{2/x}

Question
Limits and continuity
Evaluate the following limit.
$$\lim_{x\rightarrow0^+}(1+5x)^{2/x}$$

2020-12-01
Evaluate limits:
$$\lim_{x\rightarrow0^+}(1+5x)^{2/x}$$
Apply exponent rule:
$$a^x=e^{\ln(a^x)}=e^{x\cdot\ln(a)}$$
$$(1+5x)^{\frac{2}{x}}=e^{\frac{2}{x}\ln(1+5x)}$$
Apply the Limit Chain Rule:
$$g(x)=\frac{2}{x}\ln(1+5x),f(u)=e^u$$
$$\lim_{x\rightarrow0^+}g(x)=\lim_{x\rightarrow0^+}\frac{2}{x}\ln(1+5x)$$
$$=2\cdot\lim_{x\rightarrow0^+}(\frac{\ln(1+5x)}{x})$$
$$\lim_{x\rightarrow0^+}(\frac{\ln(1+5x)}{x})=\frac{0}{0}$$
L'Hospital Rule is used in the following cases,
$$\lim_{x\rightarrow a}(\frac{f(x)}{g(x)})=\frac{0}{0}\ OR\ \lim_{x\rightarrow a}(\frac{f(x)}{g(x)})=\frac{\pm\infty}{\pm\infty}$$
where a can be any real number, $$\infty\ or\ -\infty$$
then,
$$\lim_{x\rightarrow a}(\frac{f(x)}{g(x)})= \lim_{x\rightarrow a}(\frac{f'(x)}{g'(x)})$$
$$\lim_{x\rightarrow0^+}\frac{2}{x}\ln(1+5x)=2\cdot\lim_{x\rightarrow0^+}(\frac{\ln(1+5x)}{x})$$
$$=2\cdot\lim_{x\rightarrow0^+}(\frac{\frac{5}{1+5x}}{1})$$
$$=2\cdot(\frac{5}{1+5(0)})$$
$$=2\cdot5$$
$$=10$$
Result: $$\lim_{x\rightarrow0^+}(1+5x)^{2/x}=e^{10}$$

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