Find Limits of Functions in Calculus

Find the limits of various functions using different methods. Several Examples with detailed solutions are presented. More exercises with answers are at the end of this page.

Examples with Detailed Solutions

Example 1

Find the limit \[ \lim_{x \to 1^-} \dfrac{x^2 + 2x - 3}{|x-1|} \]

Solution to Example 1:

Note that we are looking for the limit as \( x \) approaches 1 from the left (\( x \to 1^- \) means \( x \) approaches 1 by values smaller than 1). Hence: \[ x \lt 1 \] \[ x - 1 \lt 0 \] If \( x - 1 \lt 0 \) then: \[ |x - 1| = -(x - 1) \] Simplify to obtain: \[ \lim_{x \to 1^-} \dfrac{(x - 1)(x + 3)}{-(x - 1)} \] \[ \lim_{x \to 1^-} \dfrac{(x - 1)(x + 3)}{-(x - 1)} = \lim_{x \to 1^-} -(x + 3) = -4 \] The final answer is: \[ \boxed{-4} \]

Example 2

Evaluate the limit \[ \lim_{x \to 5} \dfrac{x^2 - 25}{x^2 + x - 30} \]

Solution to Example 2:

Although the limit in question is the ratio of two polynomials, \( x = 5 \) makes both the numerator and denominator equal to zero. We need to factor both numerator and denominator as shown below. \[ \lim_{x \to 5} \dfrac{(x - 5)(x + 5)}{(x - 5)(x + 6)} \] Simplify to obtain: \[ \lim_{x \to 5} \dfrac{x^2 - 25}{x^2 + x - 30} = \lim_{x \to 5} \dfrac{(x + 5)}{(x + 6)} = \dfrac{10}{11} \] The final answer is: \[ \boxed{\dfrac{10}{11}} \]

Example 3

Determine the limit \[ \lim_{x \to 2} \dfrac{x^2 + 4x - 12}{|x - 2|} \]

Solution to Example 3:

We need to look at the limit from the left of 2 and the limit from the right of 2. As \( x \) approaches 2 from the left (\( x \to 2^- \)), \( x - 2 \lt 0 \), hence: \[ |x - 2| = -(x - 2) \] Substitute to obtain the limit from the left of 2 as follows: \[ \lim_{x \to 2^-} \dfrac{x^2 + 4x - 12}{-(x - 2)} \] Factor the numerator: \( x^2 + 4x - 12 = (x - 2)(x + 6) \) \[ = \lim_{x \to 2^-} \dfrac{(x - 2)(x + 6)}{-(x - 2)} = \lim_{x \to 2^-} -(x + 6) = -8 \] As \( x \) approaches 2 from the right (\( x \to 2^+ \)), \( x - 2 > 0 \), hence: \[ |x - 2| = x - 2 \] Substitute to obtain the limit from the right of 2 as follows: \[ \lim_{x \to 2^+} \dfrac{x^2 + 4x - 12}{x - 2} = \lim_{x \to 2^+} \dfrac{(x - 2)(x + 6)}{x - 2} = \lim_{x \to 2^+} (x + 6) = 8 \] The limit from the right of 2 (8) and the limit from the left of 2 (-8) are not equal, therefore the given limit DOES NOT EXIST.

Example 4

Calculate the limit \[ \lim_{x \to -1^+} \sqrt[3]{x+1} \, \ln(x+1) \]

Solution to Example 4:

As \( x \) approaches -1 from the right, \( \sqrt[3]{x+1} \) approaches 0 and \( \ln(x+1) \) approaches \( -\infty \), hence an indeterminate form \( 0 \times (-\infty) \). \[ \lim_{x \to -1^+} \sqrt[3]{x+1} \, \ln(x+1) = 0 \cdot (-\infty) \] Let us rewrite the limit so that it is of the \( \dfrac{\infty}{\infty} \) indeterminate form. \[ = \lim_{x \to -1^+} \dfrac{\ln(x+1)}{(x+1)^{-1/3}} = \dfrac{-\infty}{\infty} \] We now use L'Hôpital's Rule and find the limit. \[ = \lim_{x \to -1^+} \dfrac{\dfrac{d}{dx}[\ln(x+1)]}{\dfrac{d}{dx}[(x+1)^{-1/3}]} = \lim_{x \to -1^+} \dfrac{\dfrac{1}{x+1}}{-\dfrac{1}{3}(x+1)^{-4/3}} \] \[ = \lim_{x \to -1^+} \dfrac{1}{x+1} \cdot \left( -\dfrac{3}{(x+1)^{-4/3}} \right) = \lim_{x \to -1^+} -3 (x+1)^{1/3} = 0 \] Thus, the limit is 0. \[ \boxed{0} \]

Example 5

Find the limit: \[ \lim_{x \to \infty} - (x + 1)\left(e^{\dfrac{1}{x+1}} - 1\right) \]

Solution to Example 5:

As \( x \) gets larger, \( x + 1 \) gets larger, \( \dfrac{1}{x+1} \) approaches zero, \( e^{1/(x+1)} \) approaches 1, and \( e^{1/(x+1)} - 1 \) approaches 0; hence an indeterminate form: \( \infty \times 0 \). \[ \lim_{x \to \infty} - (x + 1)\left(e^{\dfrac{1}{x+1}} - 1\right) = -\,\infty \cdot 0 \] Let us rewrite the limit so that it is of the indeterminate form \( \dfrac{0}{0} \): \[ = \lim_{x \to \infty} -\,\dfrac{e^{\dfrac{1}{x+1}} - 1}{\dfrac{1}{x+1}} = \dfrac{0}{0} \] Apply L'Hospital's theorem to find the limit: \[ \lim_{x \to \infty} -\,\dfrac{e^{\dfrac{1}{x+1}} - 1}{\dfrac{1}{x+1}} = -\,\lim_{x \to \infty} \dfrac{\dfrac{d}{dx}\left(e^{\dfrac{1}{x+1}} - 1\right)} {\dfrac{d}{dx}\left(\dfrac{1}{x+1}\right)}. \] \[ = \lim_{x \to \infty} -\,\dfrac{\left(-\dfrac{1}{(x+1)^2}\right)e^{\dfrac{1}{x+1}}}{-\dfrac{1}{(x+1)^2}} = \lim_{x \to \infty} -\, e^{\dfrac{1}{x+1}} = -1 \] Hence \[ \lim_{x \to \infty} - (x + 1)\left(e^{\dfrac{1}{x+1}} - 1\right) = -1 \]

Example 6

Calculate the limit \[ \lim_{x \to 9} \dfrac{\sqrt{x} - 3}{x - 9} \]

Solution to Example 6:

As \( x \) approaches 9, both numerator and denominator approach 0. Multiply both numerator and denominator by the conjugate of the numerator: \[ \lim_{x \to 9} \dfrac{(\sqrt{x} - 3)(\sqrt{x} + 3)}{(x - 9)(\sqrt{x} + 3)} \] Expand and simplify: \[ \lim_{x \to 9} \dfrac{x - 9}{(x - 9)(\sqrt{x} + 3)} = \lim_{x \to 9} \dfrac{1}{\sqrt{x} + 3} \] Now find the limit: \[ \lim_{x \to 9} \dfrac{1}{\sqrt{x} + 3} = \dfrac{1}{6}. \]

Example 7

Find the limit \[ \lim_{x \to \infty} \dfrac{3 \cos x}{x} \]

Solution to Example 7:

\[ \textcolor{red}{\text{Solution to Example 7:}} \] The range of the cosine function is \[ -1 \le \cos x \le 1 \] Divide all terms of the above inequality by \( x \), for \( x > 0 \): \[ -\dfrac{1}{x} \le \dfrac{\cos x}{x} \le \dfrac{1}{x} \] Now as \( x \to +\infty \), both \( -\dfrac{1}{x} \) and \( \dfrac{1}{x} \) approach \( 0 \). Hence, by the squeezing theorem, the limit is \[ \lim_{x \to \infty} \dfrac{3 \cos x}{x} = 0. \]

Example 8

Find the limit \[ \lim_{t \to 0} \dfrac{\sin t - t}{\tan t} \]

Solution to Example 8:

As \( t \) approaches 0, both the numerator and denominator approach 0 and we have the \( 0 / 0 \) indeterminate form. \[ \lim_{t \to 0} \dfrac{\sin t - t}{\tan t} = \dfrac{0}{0} \] Hence, L'Hopital theorem is used to calculate the above limit as follows: \[ \lim_{t \to 0} \dfrac{\sin t - t}{\tan t} = \lim_{t \to 0} \dfrac{\dfrac{d}{dt}(\sin t - t)}{\dfrac{d}{dt}(\tan t)} \] \[ = \lim_{t \to 0} \dfrac{\cos t - 1}{\sec^2 t} \] \[ = \dfrac{0}{1} \] \[ = 0 \]

Example 9

Calculate the limit \[ \lim_{x \to \infty} \dfrac{3x}{\sqrt{16x^2 + 1}} \]

Solution to Example 9:

We first factor out \( 16x^2 \) under the square root of the denominator and take out of the square root and rewrite the limit as \[ = \lim_{x \to \infty} \dfrac{3x}{4|x| \sqrt{1 + \dfrac{1}{16x^2}}} \] Since \( x \) approaches larger positive values (infinity) \( |x| = x \) and \( \dfrac{1}{16x^2} \) approaches \( 0 \). Simplify and find the limit. \[ = \lim_{x \to \infty} \dfrac{3x}{4x \sqrt{1 + \dfrac{1}{16x^2}}} \] \[ = \dfrac{3}{4} \]

Example 10

Find the limit \[ \lim_{x \to 2} \dfrac{x^2 + 1}{x - 2} \]

Solution to Example 10:

As \( x \) approaches 2 from the left, then \( x - 2 \) approaches 0 from the left or \( x - 2 \lt 0 \). The numerator approaches 5 and the denominator approaches 0 from the left, hence the limit is given by \[ \lim_{x \to 2} \dfrac{x^2 + 1}{x - 2} = -\infty \]

Example 11

Calculate the limit \[ \lim_{x \to \infty} \dfrac{3x^2}{4x^2 + 2x - 1} \]

Solution to Example 11:

Factor \( x^2 \) in the denominator and simplify. \[ = \lim_{x \to \infty} \dfrac{3x^2}{x^2\left(4 + \dfrac{2}{x} - \dfrac{1}{x^2}\right)} \] As \( x \) takes large values (infinity), the terms \( \dfrac{2}{x} \) and \( \dfrac{1}{x^2} \) approach 0 hence the limit is \[ = \lim_{x \to \infty} \dfrac{3x^2}{4x^2} = \dfrac{3}{4} \]

Example 12

Compute the limit: \[ \lim_{t \to 0} \dfrac{\sin(3t)}{\sin(t)} \]

Solution to Example 12:

Multiply numerator and denominator by \( t \): \[ = \lim_{t \to 0} \dfrac{\sin(3t)}{t} \cdot \dfrac{t}{\sin(t)} \] Rewrite the expression using algebraic manipulation: \[ = \lim_{t \to 0} \dfrac{\sin(3t)}{3t} \cdot \dfrac{3t}{\sin(t)} = \lim_{t \to 0} \dfrac{\sin(3t)}{3t} \cdot 3 \cdot \dfrac{t}{\sin(t)} \] Use the known limit \( \lim_{u \to 0} \dfrac{\sin u}{u} = 1 \). Let \( u = 3t \), so as \( t \to 0 \), \( u \to 0 \): \[ \lim_{t \to 0} \dfrac{\sin(3t)}{3t} = 1 \] Also: \[ \lim_{t \to 0} \dfrac{t}{\sin(t)} = 1 \] Hence, the limit is: \[ = 1 \cdot 3 \cdot 1 = 3 \]

Example 13

Find the limit \[ \lim_{x \to \infty} \left( \sqrt{x^2 + x + 1} - x \right) \]

Solution to Example 13:

Factor \( x^2 \) inside the square root \[ \lim_{x \to \infty} \left( \sqrt{x^2 + x + 1} - x \right) = \lim_{x \to \infty} \left( \sqrt{x^2(1+\dfrac{1}{x} + \dfrac{1}{x^2}} - x ) \right) \] Use the fact that \( \sqrt{x^2} = |x| \): \[ = \lim_{x \to \infty} \left( \sqrt{x^2} \sqrt{1 + \dfrac{1}{x} + \dfrac{1}{x^2}} - x \right) = \lim_{x \to \infty} \left( |x| \sqrt{1 + \dfrac{1}{x} + \dfrac{1}{x^2}} - x \right) \] Since \( x \) takes large positive values (infinity), then \( |x| = x \) and both \( \dfrac{1}{x} \) and \( \dfrac{1}{x^2} \) approach 0 . Hence we have the indeterminate form \( \infty - \infty \): \[ = \lim_{x \to \infty} \left( x \sqrt{1 + \dfrac{1}{x} + \dfrac{1}{x^2}} - x \right) = \infty - \infty \] Restart with given limit again and multiply its numerator and denominator by the conjugate and simplify: \[ = \lim_{x \to \infty} \dfrac{ \left( \sqrt{x^2 + x + 1} - x \right) \left( \sqrt{x^2 + x + 1} + x \right) }{ \sqrt{x^2 + x + 1} + x } = \lim_{x \to \infty} \dfrac{ x^2 + x + 1 - x^2 }{ \sqrt{x^2 + x + 1} + x } = \lim_{x \to \infty} \dfrac{ x + 1 }{ \sqrt{x^2 + x + 1} + x } \] Factor \( x \) out of the numerator and denominator and simplify: \[ = \lim_{x \to \infty} \dfrac{ x \left( 1 + \dfrac{1}{x} \right) }{ x \left( \sqrt{1 + \dfrac{1}{x} + \dfrac{1}{x^2}} + 1 \right) } = \lim_{x \to \infty} \dfrac{ 1 + \dfrac{1}{x} }{ \sqrt{1 + \dfrac{1}{x} + \dfrac{1}{x^2}} + 1 } \] As \( x \) gets larger, the terms \( \dfrac{1}{x} \) and \( \dfrac{1}{x^2} \) approach zero and the limit is: \[ = \dfrac{1}{1 + 1} = \dfrac{1}{2} \]

Example 14

Determine the limit \[ \lim_{x \to \infty} x \sin \dfrac{1}{x} \]

Solution to Example 14:

Let \( z = 1 / x \) so that as \( x \) get large \( z \) approaches 0. Substitute and calculate the limit as follows. \[ = \lim_{z \to 0} \dfrac{\sin z}{z} = 1 \]

Exercises

Calculate the following limits 1) \[\lim_{x \to 2} \dfrac{x^2 - 1}{x - 1}\] 2) \[\lim_{x \to -\infty} \dfrac{3 - x}{\sqrt{x^2 + 3x}}\] 3) \[\lim_{x \to 2^+} \dfrac{x - 2}{|x - 2|}\] 4) \[\lim_{x \to 0} \dfrac{\sqrt{x + 4} - 2}{x}\] 5) \[\lim_{x \to 0} \dfrac{\tan x - x}{\sin x}\] 6) \[\lim_{x \to 0} \dfrac{\sin 4x}{\sin x}\]

Answers to Above Exercises

1) \( 3 \)
2) \( 1 \)
3) \( 1 \)
4) \( 1/4 \)
5) \( 0 \)
6) \( 4 \)