What are Common Methods to Evaluate Limits?

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There are many methods to evaluate limits. The most common methods are:

• Limits laws
• L’Hospital’s principle
• Some elementary methods, such as factoring, rationalization, compare the order of infinity, etc.

There are many other methods to evaluate limits such as squeeze theorem, the definition of definite integral, Taylor expansion, etc. But for us, those methods above are enough for use.

The limits laws are basic for everyone and we should master it. We don’t want to explain it further. Each time when we evaluate a limit, we should check if the limits laws can be applied. If it is not, we then should use another method to do it.

Now we list some cases and illustrate how to apply those methods.

• If \(f(x)=\frac{p(x)}{q(x)}\) is a rational function, where \(p(x), q(x)\) are polynomials and \(\lim_{x\to a}p(x)=0, \lim_{x\to a}q(x)=0 \), then they have the common factor \(x-a\), and we can cacel this factor and then evaluate the limit using limit laws. For example,\[\lim_{x\to1}\frac{x^2-x+2}{x^2-1}=\lim_{x\to1}\frac{(x+2)(x-1)}{(x+1)(x-1)}=\lim_{x\to1}\frac{x+2}{x+1}=\frac{3}{2}\]
• If \(f(x)\) is a rational function and both \(p(x), q(x)\) approaches to infinity, then the method of compare the order of infinity can be applied. For example,\[\lim_{x\to\infty}\frac{3x^2+2x-7}{5x^2-4x-1}=\frac{3+2/x-7/x^2}{2-4/x-1/x^2}=\frac{3}{5}.\]Here we divided both nominator and denominator by the highest order term \(x^2\). In fact, if the order of the nominator is higher than the the denominator, the result is infinity (positive or negative, depends on the direction and the order). If the order of the nominator is less than the denominator, the result is 0. if the order of the nominator and denominator equals, the result is the ratio of the coefficients of the highest terms.
• If \(f(x)\) has radical terms, it implies rationalization should be applied. For example, \begin{align*}\lim_{x\to\infty}(\sqrt{x^2+x}-\sqrt{x^2-x})&=\lim_{x\to\infty}\frac{ (\sqrt{x^2+x}-\sqrt{x^2-x}) (\sqrt{x^2+x}+\sqrt{x^2-x}) }{ \sqrt{x^2+x}+\sqrt{x^2-x} }\\ &=\lim_{x\to\infty}\frac{2x}{ \sqrt{x^2+x}+\sqrt{x^2-x} }\\ &= \lim_{x\to\infty}\frac{2}{ \sqrt{1+1/x}+\sqrt{1-1/x} } =1.\end{align*} The last second equality is attained by divide both nominator and denominator by \(x\).
• L’Hospital’s rule is the widest used method to evaluate indeterminate limits. It can deal with almost all indeterminate limits.
• For two basic indeterminate forms, \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\), this rule can be applied directly. \[\lim_{x\to0}\frac{x-\sin x}{x^3}=\lim_{x\to0}\frac{1-\cos x}{3x^2}=\lim_{x\to0}\frac{\sin x}{6x}=\lim_{x\to0}\frac{\cos x}{6}=\frac{1}{6}.\]
• For the type \(0\cdot\infty\), we change the form to be \(\frac{0}{1/\infty}\) (it becomes the form \(\frac{0}{0}\)) or \(\frac{\infty}{1/0}\) (it becomes the form \(\frac{\infty}{\infty}\)), then applied l’Hospital’s rule. \[\lim_{x\to 0^+}x^n\ln x=\lim_{x\to0^+}\frac{\ln x}{\frac{1}{x^n}}=\lim_{x\to0^+}\frac{\frac{1}{x}}{-\frac{n}{x^{n+1}}}=\lim_{x\to0^+}\frac{x^{n+1}}{-nx}=0\]
• For the type \(\infty-\infty\), we use common denominator or other algebraic operation to transform it to the basic form. \[\lim_{x\to\frac{\pi}{2}}(\sec x-\tan x)=\lim_{x\to\frac{\pi}{2}}\frac{1-\sin x}{\cos x}=\lim_{x\to\frac{\pi}{2}}\frac{-\cos x}{-\sin x}=0\]
• For the types \(0^0, \infty^0, 1^{\infty}\), we use the identity \(x=e^{\ln x}\) to transform the form to be the basic form. \[\lim_{x\to0^+}x^x=\lim_{x\to0^+}e^{\ln(x^x)}= \lim_{x\to0^+}e^{x\ln x}=e^{ \lim_{x\to0^+}x\ln x}. \]Since \[ \lim_{x\to0^+}x\ln x = \lim_{x\to0^+}\frac{\ln x}{\frac{1}{x}}= \lim_{x\to0^+}\frac{\frac{1}{ x}}{-\frac{1}{x^2}}= \lim_{x\to0^+}-x=0. \] So the original limit equals \(e^0=1\).