Limits by Rationalization

We have seen several methods for finding limits, including limits by substitution, limits by factoring, and using the epsilon-delta definition of the limit.

In the case when direct substitution into the function gives an indeterminate form $\big($such as $\frac{0}{0}$ or $\frac{\infty}{\infty}\big)$ and the function involves a radical expression or a trigonometric function, it may be possible to find the limit by multiplying by a conjugate.

If the function in the limit involves a square root or a trigonometric function, it may be possible to simplify the expression by multiplying by the conjugate. This method uses the algebraic identity

$$( x - y )( x + y ) = x^2-y^2.$$

For example, given the expression $\sqrt{a} - \sqrt{b},$ the conjugate is $\sqrt{a} + \sqrt{b},$ and multiplying by $\frac{ \sqrt{a} + \sqrt{b} }{ \sqrt{a} + \sqrt{b} }$ gives

$$\begin{aligned} \big( \sqrt{a} - \sqrt{b} \big) \cdot \frac{ \sqrt{a} + \sqrt{b} }{ \sqrt{a} + \sqrt{b} } &= \frac{ \big(\sqrt{a}\big)^2 - \big(\sqrt{b}\big)^2}{\sqrt{a} + \sqrt{b} } \\ & = \frac{a - b}{\sqrt{a} + \sqrt{b} }. \end{aligned}$$

In the case the function involves a trigonometric expression, it may be possible to simplify after conjugation by applying a trigonometric identity.

Find the limit

$$\lim_{x \rightarrow 0} \frac{ \sqrt{1+x} - 1}{x}.$$

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Note that direct substitution gives the indeterminate form $\frac{0}{0}$. We multiply both the numerator and denominator by the conjugate of the numerator $\sqrt{1+x} + 1 :$

$$\begin{aligned} \lim_{x \rightarrow 0} \frac{ \sqrt{1+x} - 1}{x} & = \lim_{x \rightarrow 0} \frac{ \big( \sqrt{1+x} - 1\big) \big( \sqrt{1+x} + 1\big) }{x \big( \sqrt{1+x} + 1\big) } \\\\ &= \lim_{x \rightarrow 0} \frac{ (1 + x) - 1 }{x \big( \sqrt{1+x} + 1\big) } \\\\ &= \lim_{x \rightarrow 0} \frac{ x }{x \big( \sqrt{1+x} + 1\big) } \\\\ &= \frac{1}{2}.\ _\square \end{aligned}$$

Calculate the derivative of the function $f(x) = \sqrt{x}$ from first principles.

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By the definition of a derivative, we have

$$\begin{aligned} f'(x) & = \lim_{h \rightarrow 0 } \frac{ f (x + h) - f(x) }{h} \\\\ &=\lim_{h \rightarrow 0 } \frac{ \sqrt{x + h} - \sqrt{x} }{h} \\\\ &=\lim_{h \rightarrow 0 } \frac{ \big( \sqrt{x + h} - \sqrt{x} \big)\big( \sqrt{x + h} + \sqrt{x} \big) }{h \big( \sqrt{x + h} + \sqrt{x} \big) } \\\\ &= \lim_{h \rightarrow 0 } \frac{ ( x + h) - x }{h \big( \sqrt{x + h} + \sqrt{x} \big) } \\\\ &= \lim_{h \rightarrow 0 } \frac{ 1 }{ \sqrt{x + h} + \sqrt{x} }\\ &= \frac{1}{ 2 \sqrt{x} }.\ _\square \end{aligned}$$

Find the limit

$$\lim_{x \rightarrow 0} \frac{\sec x - 1}{x^2}.$$

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In this limit, direct substitution gives the indeterminate form $\frac{0}{0}$. We multiply both the numerator and denominator by the conjugate $\sec x + 1$ of the numerator. This gives

$$\begin{aligned} \lim_{x \rightarrow 0} \frac{\sec x - 1}{x^2} &= \lim_{x \rightarrow 0} \frac{(\sec x - 1)(\sec x + 1)}{x^2 (\sec x + 1)} \\\\ &= \lim_{x \rightarrow 0} \frac{\tan^2 x}{x^2 (\sec x + 1) } \\\\ & =\lim_{x \rightarrow 0} \left( \frac{\sin x}{x} \right)^2 \left( \frac{1}{\cos^2 x} \right) \left( \frac{1}{\sec x + 1} \right) \\\\ & = 1^2 \cdot \frac{1}{1} \cdot \frac{1}{2} = \frac{1}{2}, \end{aligned}$$

where the limit $\lim_{x \rightarrow 0} \frac{\sin x}{x}$ was calculated in epsilon-delta definition of a limit. $_\square$

If the radical expression involves cube or higher roots, it may be possible to use the algebraic identity

$$x^n-y^n = (x-y)(x^{n-1} + x^{n-2} y + \ldots + xy^{n-2} + y^{n-1} ).$$

to simplify the expression.

Find the limit

$$\lim_{x \rightarrow 0} \frac{ 1 - \sqrt[3]{x^2 + 1} }{x^2}.$$

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In this limit, direct substitution gives the indeterminate form $\frac{0}{0}$. We multiply both the numerator and denominator by $\left( 1 + \sqrt[3]{x^2 + 1} + \left( \sqrt[3]{x^2 + 1} \right)^2 \right)$ to simplify the numerator. This gives

$$\begin{aligned} \lim_{x \rightarrow 0} \frac{ 1 - \sqrt[3]{x^2 + 1} }{x^2} &= \lim_{x \rightarrow 0} \frac{ 1 - \sqrt[3]{x^2 + 1} }{x^2} \frac{ \left( 1 + \sqrt[3]{x^2 + 1} + \left( \sqrt[3]{x^2 + 1} \right)^2 \right) }{ \left( 1 + \sqrt[3]{x^2 + 1} + \left( \sqrt[3]{x^2 + 1} \right)^2 \right) }\\ &= \lim_{x \rightarrow 0} \frac{ 1 - (x^2 + 1) }{ x^2 \left( 1 + \sqrt[3]{x^2 + 1} + \left( \sqrt[3]{x^2 + 1} \right)^2 \right) }\\ &= \lim_{x \rightarrow 0} \frac{-x^2}{x^2 \left( 1 + \sqrt[3]{x^2 + 1} + \left( \sqrt[3]{x^2 + 1} \right)^2 \right) }\\ &= \lim_{x \rightarrow 0} \frac{-1}{\left( 1 + \sqrt[3]{x^2 + 1} + \left( \sqrt[3]{x^2 + 1} \right)^2 \right) }\\ &=\frac{-1}{\left( 1 + \sqrt[3]{1} + \left( \sqrt[3]{1} \right)^2 \right) }\\ &= -\frac{1}{3}.\ _\square \end{aligned}$$