Find the limits of various functions using different methods, including factoring, conjugate multiplication, L'Hôpital's Rule, and the Squeezing Theorem. Examples with detailed step-by-step solutions are presented below.
In what follows, \( C \) represents the constant of integration where applicable.
Examples with Detailed Solutions
Click on each example to view its detailed step-by-step solution.
Example 1
Find the limit:
\[ \lim_{x \to 1^-} \dfrac{x^2 + 2x - 3}{|x-1|} \]Show Solution
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:
\[ x < 1 \] \[ x - 1 < 0 \]If \( x - 1 < 0 \) then:
\[ |x - 1| = -(x - 1) \]Factor the numerator and 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:
\[ \lim_{x \to 5} \dfrac{x^2 - 25}{x^2 + x - 30} \]Show Solution
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 the 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:
\[ \lim_{x \to 2} \dfrac{x^2 + 4x - 12}{|x - 2|} \]Show Solution
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 (\( x \to 2^- \)), \( x - 2 < 0 \), 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 two-sided limit DOES NOT EXIST.
Example 4
Calculate the limit:
\[ \lim_{x \to -1^+} \sqrt[3]{x+1} \, \ln(x+1) \]Show Solution
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:
\[ \lim_{x \to \infty} - (x + 1)\left(e^{\dfrac{1}{x+1}} - 1\right) \]Show Solution
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'Hôpital'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:
\[ \lim_{x \to 9} \dfrac{\sqrt{x} - 3}{x - 9} \]Show Solution
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 by direct substitution:
\[ \lim_{x \to 9} \dfrac{1}{\sqrt{x} + 3} = \dfrac{1}{6} \]Example 7
Find the limit:
\[ \lim_{x \to \infty} \dfrac{3 \cos x}{x} \]Show Solution
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:
\[ \lim_{t \to 0} \dfrac{\sin t - t}{\tan t} \]Show Solution
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'Hôpital's 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:
\[ \lim_{x \to \infty} \dfrac{3x}{\sqrt{16x^2 + 1}} \]Show Solution
We first factor out \( 16x^2 \) under the square root of the denominator, take it 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:
\[ \lim_{x \to 2^-} \dfrac{x^2 + 1}{x - 2} \]Show Solution
As \( x \) approaches 2 from the left, then \( x - 2 \) approaches 0 from the left or \( x - 2 < 0 \). 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:
\[ \lim_{x \to \infty} \dfrac{3x^2}{4x^2 + 2x - 1} \]Show Solution
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:
\[ \lim_{t \to 0} \dfrac{\sin(3t)}{\sin(t)} \]Show Solution
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:
\[ \lim_{x \to \infty} \left( \sqrt{x^2 + x + 1} - x \right) \]Show Solution
Factor \( x^2 \) inside the square root:
\[ \lim_{x \to \infty} \left( \sqrt{x^2 + x + 1} - x \right) = \lim_{x \to \infty} \left( \sqrt{x^2\left(1+\dfrac{1}{x} + \dfrac{1}{x^2}\right)} - x \right) \]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 the given limit again, 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 evaluates to:
\[ = \dfrac{1}{1 + 1} = \dfrac{1}{2} \]Example 14
Determine the limit:
\[ \lim_{x \to \infty} x \sin \dfrac{1}{x} \]Show Solution
Let \( z = 1 / x \) so that as \( x \) gets large, \( z \) approaches 0. Substitute and calculate the limit as follows:
\[ = \lim_{z \to 0} \dfrac{\sin z}{z} = 1 \]Practice Exercises
Calculate the following limits. Click each exercise to check your answers.
Exercise 1
\( \displaystyle \lim_{x \to 2} \dfrac{x^2 - 1}{x - 1} \)
Show Answer
Exercise 2
\( \displaystyle \lim_{x \to -\infty} \dfrac{3 - x}{\sqrt{x^2 + 3x}} \)
Show Answer
Exercise 3
\( \displaystyle \lim_{x \to 2^+} \dfrac{x - 2}{|x - 2|} \)
Show Answer
Exercise 4
\( \displaystyle \lim_{x \to 0} \dfrac{\sqrt{x + 4} - 2}{x} \)
Show Answer
Exercise 5
\( \displaystyle \lim_{x \to 0} \dfrac{\tan x - x}{\sin x} \)
Show Answer
Exercise 6
\( \displaystyle \lim_{x \to 0} \dfrac{\sin 4x}{\sin x} \)