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.
Multiplying by the 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
\[\begin{align} \big( \sqrt{a} - \sqrt{b} \big) \cdot \frac{ \sqrt{a} + \sqrt{b} }{ \sqrt{a} + \sqrt{b} } &= \frac{ \big(\sqrt{a}\big)^2 - \big(\sqrt{b}\big)^2}{\sqrt{a} + \sqrt{b} } \\ & = \frac{a - b}{\sqrt{a} + \sqrt{b} }. \end{align}\]
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
\[ \lim_{x \rightarrow 0} \frac{ \sqrt{1+x} - 1}{x}. \]
Note that direct substitution gives the indeterminate form \(\frac{0}{0}\). We multiply both the numerator and denominator by the conjugate of the numerator \( \sqrt{1+x} + 1 :\)
\[ \begin{align} \lim_{x \rightarrow 0} \frac{ \sqrt{1+x} - 1}{x} & = \lim_{x \rightarrow 0} \frac{ \big( \sqrt{1+x} - 1\big) \big( \sqrt{1+x} + 1\big) }{x \big( \sqrt{1+x} + 1\big) } \\\\ &= \lim_{x \rightarrow 0} \frac{ (1 + x) - 1 }{x \big( \sqrt{1+x} + 1\big) } \\\\ &= \lim_{x \rightarrow 0} \frac{ x }{x \big( \sqrt{1+x} + 1\big) } \\\\ &= \frac{1}{2}.\ _\square \end{align}\]
Calculate the derivative of the function \(f(x) = \sqrt{x}\) from first principles.
By the definition of a derivative, we have
\[ \begin{align} f'(x) & = \lim_{h \rightarrow 0 } \frac{ f (x + h) - f(x) }{h} \\\\ &=\lim_{h \rightarrow 0 } \frac{ \sqrt{x + h} - \sqrt{x} }{h} \\\\ &=\lim_{h \rightarrow 0 } \frac{ \big( \sqrt{x + h} - \sqrt{x} \big)\big( \sqrt{x + h} + \sqrt{x} \big) }{h \big( \sqrt{x + h} + \sqrt{x} \big) } \\\\ &= \lim_{h \rightarrow 0 } \frac{ ( x + h) - x }{h \big( \sqrt{x + h} + \sqrt{x} \big) } \\\\ &= \lim_{h \rightarrow 0 } \frac{ 1 }{ \sqrt{x + h} + \sqrt{x} }\\ &= \frac{1}{ 2 \sqrt{x} }.\ _\square \end{align}\]
Find the limit
\[ \lim_{x \rightarrow 0} \frac{\sec x - 1}{x^2}.\]
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
\[ \begin{align} \lim_{x \rightarrow 0} \frac{\sec x - 1}{x^2} &= \lim_{x \rightarrow 0} \frac{(\sec x - 1)(\sec x + 1)}{x^2 (\sec x + 1)} \\\\ &= \lim_{x \rightarrow 0} \frac{\tan^2 x}{x^2 (\sec x + 1) } \\\\ & =\lim_{x \rightarrow 0} \left( \frac{\sin x}{x} \right)^2 \left( \frac{1}{\cos^2 x} \right) \left( \frac{1}{\sec x + 1} \right) \\\\ & = 1^2 \cdot \frac{1}{1} \cdot \frac{1}{2} = \frac{1}{2}, \end{align} \]
where the limit \(\lim_{x \rightarrow 0} \frac{\sin x}{x} \) was calculated in epsilon-delta definition of a limit. \(_\square\)
Other Radical Roots
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
\[ \lim_{x \rightarrow 0} \frac{ 1 - \sqrt[3]{x^2 + 1} }{x^2}.\]
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
\[ \begin{align} \lim_{x \rightarrow 0} \frac{ 1 - \sqrt[3]{x^2 + 1} }{x^2} &= \lim_{x \rightarrow 0} \frac{ 1 - \sqrt[3]{x^2 + 1} }{x^2} \frac{ \left( 1 + \sqrt[3]{x^2 + 1} + \left( \sqrt[3]{x^2 + 1} \right)^2 \right) }{ \left( 1 + \sqrt[3]{x^2 + 1} + \left( \sqrt[3]{x^2 + 1} \right)^2 \right) }\\ &= \lim_{x \rightarrow 0} \frac{ 1 - (x^2 + 1) }{ x^2 \left( 1 + \sqrt[3]{x^2 + 1} + \left( \sqrt[3]{x^2 + 1} \right)^2 \right) }\\ &= \lim_{x \rightarrow 0} \frac{-x^2}{x^2 \left( 1 + \sqrt[3]{x^2 + 1} + \left( \sqrt[3]{x^2 + 1} \right)^2 \right) }\\ &= \lim_{x \rightarrow 0} \frac{-1}{\left( 1 + \sqrt[3]{x^2 + 1} + \left( \sqrt[3]{x^2 + 1} \right)^2 \right) }\\ &=\frac{-1}{\left( 1 + \sqrt[3]{1} + \left( \sqrt[3]{1} \right)^2 \right) }\\ &= -\frac{1}{3}.\ _\square \end{align} \]