Trigonometric Functions

In the limit of the indeterminate form of trigonometric functions, the principle is to reduce it to the following formula.

The limit \(\displaystyle\lim_{x\rightarrow 0}\frac{\sin x}{x}=1\) is not obvious. To prove the above equation, we use a geometric argument. See also proof.

Example

(1)\( \displaystyle\lim_{x\rightarrow 0}\frac{\sin 3x}{5x}=\displaystyle\lim_{x\rightarrow 0}\frac{\sin 3x}{3x}\cdot \frac{3}{5}=\frac{3}{5}\)

(2)\(\displaystyle\lim_{x\rightarrow 0} \frac{1-\cos x}{x^{2}}=\displaystyle\lim_{x\rightarrow 0} \frac{(1-\cos x)(1+\cos x)}{x^{2}(1+\cos x)}=\displaystyle\lim_{x\rightarrow 0} \frac{\sin^{2} x}{x^{2}(1+\cos x)}=\displaystyle\lim_{x\rightarrow 0} \left(\frac{\sin x}{x}\right)^{2}\cdot \frac{1}{1+\cos x}=1^{2}\cdot \frac{1}{2}=\frac{1}{2}\)

(3)\(\displaystyle\lim_{x\rightarrow 0}\frac{\tan x}{x}=\displaystyle\lim_{x\rightarrow 0}\frac{\sin x}{\cos x}\cdot \frac{1}{x}=\frac{\sin x}{x}\cdot \frac{1}{\cos x}=1\cdot \frac{1}{1}=1\)

These expression : \(\displaystyle\lim_{x\rightarrow 0}\frac{\sin bx}{ax}=\frac{b}{a}, \ \ \ \displaystyle\lim_{x\rightarrow 0} \frac{1-\cos x}{x^{2}}=\frac{1}{2}, \ \ \ \displaystyle\lim_{\theta\rightarrow 0}\frac{\tan \theta}{\theta}=1\) are worth to memorize as formulas.

Logarithmic and Exponential Functions

In the limit of the indeterminate form involving logarithmic functions or \((1+f(x))^{g(x)}\), the principle is to reduce it to the following formula.

This is the definition of the number \(\mathrm{e}\). This expression is equivalent to the following:

$$\displaystyle\lim_{x\rightarrow \pm \infty}(1+\frac{1}{x})^{x}=\mathrm{e}$$

From the limit, we can obtain the following formulas.

We prove \(\displaystyle\lim_{x\rightarrow 0}\frac{a^{x}-1}{x}=\log_{\mathrm{e}} a\). （ This is the derivative of \(f(x)=a^{x}\) at \(x = 0\); \(f'(0)\) )

Proof : If we set \(a^{x}-1=t \), then we have \(t\rightarrow 0\) when\(x\rightarrow 0\).

$$\displaystyle\lim_{x\rightarrow 0}\frac{a^{x}-1}{x}=\displaystyle\lim_{x\rightarrow 0}\frac{t (\log_{e} a)}{\log_{e} (1+t)}=\displaystyle\lim_{x\rightarrow 0}\frac{(\log_{e} a)}{\log_{e} (1+t)^{\frac{1}{t}}}=\frac{\log_{e} a}{\log_{e} e}=\log_{e} a$$

Similarly, \(\displaystyle\lim_{x\rightarrow 0}\frac{\mathrm{e}^{x}-1}{x}=1\) can be obetained by setting \(e^{x}-1=t \).

Example

(1) \(\displaystyle\lim_{x\rightarrow \infty} \left(1-\frac{1}{x^{2}}\right)^{x}=\displaystyle\lim_{x\rightarrow 0}\left(\left(1+\frac{1}{(-x^{2})}\right)^{-x^{2}}\right)^{\frac{1}{(-x)}}=\mathrm{e}^{0}=1\)

(2) \(\displaystyle\lim_{x\rightarrow 0}\frac{5^{x}-3^{x}}{x}=\displaystyle\lim_{x\rightarrow 0}\frac{(5^{x}-1)-(3^{x}-1)}{x}=\log_{e} 5-\log_{e}3=\log_{3}\frac{5}{3}\)

(3) \(\displaystyle\lim_{x\rightarrow 0}\frac{\log_{e} (1+x)}{x}=\displaystyle\lim_{x\rightarrow 0} \log_{e} (1+x)^{\frac{1}{x}}=\log_{e} e=1\)

Tricia

The following limit values can be obtained from the derivative at x = 0.

Limit Values	Derivation Method
\(\displaystyle\lim_{x\rightarrow 0}\frac{\mathrm{e}^{x}-1}{x}=1\)	\((\mathrm{e}^{x})’_{x=0}=\displaystyle\lim_{x\rightarrow 0}\frac{\mathrm{e}^{x}-\mathrm{e}^{0}}{x-0}=\displaystyle\lim_{x\rightarrow 0}\frac{\mathrm{e}^{x}-1}{x}=[\mathrm{e}^{x}]_{x=0}1\)
\(\displaystyle\lim_{x\rightarrow 0}\frac{\log (1+x)}{x}=1\)	\(\left(\log (1+x)\right)’_{x=0}=\displaystyle\lim_{x\rightarrow 0}\frac{\log (1+x)}{x}=[{\frac{1}{1+x}}]_{x=0}=1\)
\(\displaystyle\lim_{x\rightarrow 0}\frac{\sin x}{x}=1\)	\((\sin x)’_{x=0}=\displaystyle\lim_{x\rightarrow 0}\frac{\sin x}{x}=[\cos x]_{x=0}=1\)
\(\displaystyle\lim_{x\rightarrow 0}\frac{1-\cos x}{x}=0\)	\((\cos x)’_{x=0}=\displaystyle\lim_{x\rightarrow 0}\frac{\cos x-1}{x-0}=\displaystyle\lim_{x\rightarrow 0}-\left(\frac{1-\cos x}{x}\right)=[-\sin x]_{x=0}=0\)
\(\displaystyle\lim_{x\rightarrow 0}\frac{\tan x}{x}=1\)	\((\tan x)’=\displaystyle\lim_{x\rightarrow 0}\frac{\tan x-0}{x-0}=[\frac{1}{\cos^{2} x}]_{x=0}=1\)