
How do you find one-sided limits algebraically?
Answer
542.7k+ views
Hint: A one-sided limit is either of the two limits of a function f(x) of a real variable x as x approaches a specified point either from the left or from the right. When you have a left-hand limit or right-hand limit or both you can solve one-sided limits algebraically.
Complete step by step answer:
Let’s answer this question using examples.
Example 1 : $\mathop {\lim }\limits_{x \to 1} ({x^3} - 2)$
In questions like this, you have to just put the value of x
$ \Rightarrow ({1^3}) - 2$
$ \Rightarrow 1 - 2$
$ \Rightarrow - 1$
Let’s move on to the next example,
Example 2 : $\mathop {\lim }\limits_{x \to 1} \dfrac{{({x^2} - 9)}}{{x - 9}}$
Directly putting the value $x = 1$
We get $\dfrac{0}{0}$ , this $\dfrac{0}{0}$ is known as indeterminate .
In questions like this when you directly put the value you get the value in $\dfrac{0}{0}$ form.
To solve this, we should simplify the question
You can apply the formula $(a - b)(a + b) = {a^2} - {b^2}$
$ \Rightarrow \mathop {\lim }\limits_{x \to 1} \dfrac{{(x - 3)(x + 3)}}{{(x - 3)}}$
Numerator $(x - 3)$ and denominator $(x - 3)$ gets cancelled out
$ \Rightarrow \mathop {\lim }\limits_{x \to 1} (x + 3)$
$ \Rightarrow (1 + 3)$
$ \Rightarrow 4$
Additional information:
You can use L'Hospital's Rule to solve limits that are indeterminate such as $\dfrac{0}{0}$ and $\dfrac{\infty }{\infty }$.
Note:
One-sided limits are the same as normal limits, just restriction of x takes place so that it approaches from just one side.
\[x \to {a^ + }\;\]simply means x is approaching from the right side . Similarly , \[x \to {a^ - }\;\] simply means x is approaching from the left side. When you have a graph of a particular function whose one-sided limit you want to calculate, graphs make it easier to calculate .
Limits exist when the right hand limit is equal to the left hand limit.
Complete step by step answer:
Let’s answer this question using examples.
Example 1 : $\mathop {\lim }\limits_{x \to 1} ({x^3} - 2)$
In questions like this, you have to just put the value of x
$ \Rightarrow ({1^3}) - 2$
$ \Rightarrow 1 - 2$
$ \Rightarrow - 1$
Let’s move on to the next example,
Example 2 : $\mathop {\lim }\limits_{x \to 1} \dfrac{{({x^2} - 9)}}{{x - 9}}$
Directly putting the value $x = 1$
We get $\dfrac{0}{0}$ , this $\dfrac{0}{0}$ is known as indeterminate .
In questions like this when you directly put the value you get the value in $\dfrac{0}{0}$ form.
To solve this, we should simplify the question
You can apply the formula $(a - b)(a + b) = {a^2} - {b^2}$
$ \Rightarrow \mathop {\lim }\limits_{x \to 1} \dfrac{{(x - 3)(x + 3)}}{{(x - 3)}}$
Numerator $(x - 3)$ and denominator $(x - 3)$ gets cancelled out
$ \Rightarrow \mathop {\lim }\limits_{x \to 1} (x + 3)$
$ \Rightarrow (1 + 3)$
$ \Rightarrow 4$
Additional information:
You can use L'Hospital's Rule to solve limits that are indeterminate such as $\dfrac{0}{0}$ and $\dfrac{\infty }{\infty }$.
Note:
One-sided limits are the same as normal limits, just restriction of x takes place so that it approaches from just one side.
\[x \to {a^ + }\;\]simply means x is approaching from the right side . Similarly , \[x \to {a^ - }\;\] simply means x is approaching from the left side. When you have a graph of a particular function whose one-sided limit you want to calculate, graphs make it easier to calculate .
Limits exist when the right hand limit is equal to the left hand limit.
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