Chapters
In this article, we will learn how to calculate the limits of a logarithmic function in detail.
There are two primary properties of limits that are used while computing the limits of the logarithmic functions. The standard results which we get from these properties can be employed as formulas in calculus for dealing with the functions which involve the logarithmic functions. These two properties are discussed here in detail:
1) The limit of the quotient of the natural logarithm of 1 + x divided by x is equal to 1. Mathematically, we can write it as:
\lim_{x \rightarrow 0} \frac {log_e (1 +x)} {x} = 1
2) If we have the ratio of the logarithm of 1 + x to the base x, then it is equal to the reciprocal of natural logarithm of the base.
\lim_{x \rightarrow 0} \frac {log_b (1 +x)} {x} = \frac {1} {log_e b}
Now, we will learn how to evaluate the problems involving the limit of the logarithmic function.
Example 1
Evaluate \lim _ {x \rightarrow 0} \frac {ln (sin x)} {\sqrt [3] {1 + 2x} - 1}
Solution
In the first step, we will substitute x = 0 in the above function as shown below:
=\lim _ {x \rightarrow 0} \frac {ln (sin (0))} {\sqrt [3] {1 + 2(0)} - 1}
According to the trigonometry, the value of sine zero is equal to one:
=\lim _ {x \rightarrow 0} \frac {ln (1)} {\sqrt [3] {1 + 0} - 1}
Solve the remaining function arithmetically as shown below:
=\lim _ {x \rightarrow 0} \frac {0} {\sqrt [3] {1 } - 1}
=\lim _ {x \rightarrow 0} \frac {0} {0}
The result of the function in this example is an indeterminate form.
Example 2
Evaluate \lim _ {x \rightarrow 0} \frac {tan x + 2} {ln (\sqrt {1})}
Solution
In the first step, we will substitute x = 0 in the above function as shown below:
=\lim _ {x \rightarrow 0} \frac {tan (0) + 2} {ln (\sqrt {1})}
According to the trigonometry, the value of tangent zero is equal to zero:
=\lim _ {x \rightarrow 0} \frac {0 + 2} {ln (\sqrt{1}}
Solve the remaining function arithmetically as shown below:
=\lim _ {x \rightarrow 0} \frac {2} {ln (1)}
=\lim _ {x \rightarrow 0} \frac {2} {0}
The result of the function in this example is indeterminate form because 0 is in the denominator.
Example 3
Evaluate \lim _{x \rightarrow e} \frac{ln (x) - 2} {x - 2e}
Solution
Here, we will employ L'Hôpital's rule which says that if we have the function of the form 
which meet any of the following two conditions:
- Both the functions in the numerator and the denominator f(x) and g(x) have a of limit zero as x approaches a
- Both the functions in the ratio f(x) and g(x) have infinite limit positive or negative) as x approaches a
Then, the limit of the ratio of f and g, i.e. 
is equal to the limit of the ratio of derivatives of f and g , i.e., 
, until the limit exists, or is infinite.
Using this rule, we need to find the derivative of the numerator and the denominator like this:
=\lim _{x \rightarrow e} \frac{\frac{1}{x} - 0} {1 - 0}
=\lim _{x \rightarrow e} \frac{1}{x}
Now, we will substitute x in the above function with e as shown below:
=\frac{1} {e}
Example 4
Evaluate \lim _ {x \rightarrow 1} \frac {log(x)}{2x - 2}
Solution
When we substitute 1 directly in the above expression we get the indeterminate form of 
. Therefore, we will apply L'Hôpital's rule here. To use this rule, we need to take the derivative of the functions in the numerator and the denominator as shown below:
= \lim _ {x \rightarrow 1} \frac {\frac{1}{x}}{2 - 0}
= \lim _ {x \rightarrow 1} \frac {\frac{1}{x}}{2}
Now, we will substitute x with 1 in the above expression to get the final answer as shown below:
= \lim _ {x \rightarrow 0} \frac {\frac{1}{1}}{2}
Example 5
Evaluate \lim _ {x \rightarrow 3} \frac {log(x - 2)}{3x - 9}
Solution
When we substitute 3 directly in the above expression we get the indeterminate form . Therefore, we will apply L'Hôpital's rule here. To use this rule, we need to take the derivative of the functions in the numerator and the denominator as shown below:
= \lim _ {x \rightarrow 3} \frac {\frac{1}{x - 2}}{3 - 0}
= \lim _ {x \rightarrow 3} \frac {\frac{1}{x - 2}}{3}
Now, we will substitute x with 3 in the above expression to get the final answer as shown below:
= \lim _ {x \rightarrow 0} \frac {\frac{1}{1}}{3}
Example 6
Evaluate \lim _ {x \rightarrow 0} \frac {cos x + 1} {log (\sqrt {4})}
Solution
In the first step, we will substitute x = 0 in the above function as shown below:
=\lim _ {x \rightarrow 0} \frac {cos (0) + 1} {log (\sqrt {4})}
According to the trigonometry, the value of cosine zero is equal to one:
=\lim _ {x \rightarrow 0} \frac {1 + 1} {log (\sqrt{4}}
Solve the remaining function arithmetically as shown below:
=\lim _ {x \rightarrow 0} \frac {2} {log (2)}
=\lim _ {x \rightarrow 0} \frac {2} {0.301}
Example 7
Evaluate \lim _ {x \rightarrow 0} \frac {(tan x + sin x) + 3} {log (\sqrt {9})}
Solution
In the first step, we will substitute x = 0 in the above function as shown below:
\lim _ {x \rightarrow 0} \frac {(tan (0) + sin (0)) + 3} {log (\sqrt {9})}
According to the trigonometry, the values of sin and tangent zero are equal to one:
=\lim _ {x \rightarrow 0} \frac {0 + 0 + 3} {log (\sqrt{9}}
Solve the remaining function arithmetically as shown below:
=\lim _ {x \rightarrow 0} \frac {3} {log (3)}
=\lim _ {x \rightarrow 0} \frac {3} {0.301}
Example 8
Evaluate \lim _ {x \rightarrow 0} \frac {log x} {2sin x + 5}
Solution
In the first step, we will substitute x = 0 in the above function as shown below:
\lim _ {x \rightarrow 0} \frac {log (0)} {2sin (0) + 5}
According to the trigonometry, the value of sine zero is equal to one:
=\lim _ {x \rightarrow 0} \frac {log (0)} {5}
Solve the remaining function arithmetically as shown below:
=\lim _ {x \rightarrow 0} \frac {-\infty} {5}
Example 9
Evaluate \lim _ {x \rightarrow 1} \frac {log (2x -2)} {8x + 3}
Solution
In the first step, we will substitute x = 1 in the above function as shown below:
\lim _ {x \rightarrow 1} \frac {log (2(1) -2)} {8(1) + 3}
According to the trigonometry, the value of sine zero is equal to one:
=\lim _ {x \rightarrow 0} \frac {log (0)} {11}
Solve the remaining function arithmetically as shown below:
=\lim _ {x \rightarrow 0} \frac {-\infty} {11}
Example 10
Evaluate \lim _ {x \rightarrow 0} \frac {cos x + 2} {log (\sqrt {9})}
Solution
In the first step, we will substitute x = 0 in the above function as shown below:
=\lim _ {x \rightarrow 0} \frac {cos (0) + 2} {log (\sqrt {9})}
According to the trigonometry, the value of cosine zero is equal to one:
=\lim _ {x \rightarrow 0} \frac {1 + 2} {log (\sqrt{9}}
Solve the remaining function arithmetically as shown below:
=\lim _ {x \rightarrow 0} \frac {3} {log (3)}
=\lim _ {x \rightarrow 0} \frac {3} {0.477}
Example 11
Evaluate \lim _ {x \rightarrow 0} \frac {cos x + tan x - 8 } {log (\sqrt {25})}
Solution
In the first step, we will substitute x = 0 in the above function as shown below:
=\lim _ {x \rightarrow 0} \frac {cos (0) + tan (0) - 8} {log (\sqrt {25})}
According to the trigonometry, the value of cosine zero is 1 and tangent zero is zero:
=\lim _ {x \rightarrow 0} \frac {-7} {log (\sqrt{25}}
Solve the remaining function arithmetically as shown below:
=\lim _ {x \rightarrow 0} \frac {-7} {log (5)}
=\lim _ {x \rightarrow 0} \frac {-7} {0.698}
Did you like this article? Rate it!
Loading...
Derivative of denominator is wrong
Good catch—thanks for pointing that out! We’ll double-check the derivative in that section and make any necessary corrections. Really appreciate you taking the time to flag it. 👍
Thank you
Thank you Abbas! Good luck with your studies!
With regard to the Zero Over a Number item, is there a mis-statement? It’s immediately followed by “If a number is divided by zero which means that the numerator is zero and the denominator is the number, then the result is zero.”
Hi Mark,
You’re absolutely right to raise the question — there does appear to be a misstatement in that sentence. The phrase “If a number is divided by zero, which means that the numerator is zero and the denominator is the number…” is indeed misleading and should be corrected.
To clarify:
Zero divided by a number (e.g. 0 ÷ 5) equals 0.
A number divided by zero (e.g. 5 ÷ 0) is undefined.
We’ll update the sentence to reflect the correct mathematical explanation. We appreciate you catching that and helping us improve the accuracy of the content!
There is more than one size of infinity, though. What if you multiply the infinity of the whole numbers (Aleph-0) by the infinity of the real numbers (fraktur-c)?
Thanks a lot to you for this essentiol article.
Hi Piyash! Thanks for your comment, great to hear that you found this useful!
Very nice