
How do I use the properties of limit to evaluate a limit?
Answer
515.7k+ views
Hint: In the given question we have been asked to describe the properties of limit function and how to use the same to evaluate a limit. For this, first recall all the properties of limits and then give an example of how to solve a limit question using properties.
Now let \[{x_0}\] belong to real numbers set and suppose that \[\mathop {\lim }\limits_{x \to {x_0}} f(x)\] and \[\mathop {\lim }\limits_{x \to {x_0}} f(x)\] exist and are finite. If \[c \in \] real number set is a constant,
These are some important properties of limit:
\[
1)\mathop {\lim }\limits_{x \to {x_0}} [cf(x)] = c\left[ {\mathop {\lim }\limits_{x \to {x_0}} f(x)} \right] \\
2)\mathop {\lim }\limits_{x \to {x_0}} [f(x) \pm g(x)] = \left[ {\mathop {\lim }\limits_{x \to {x_0}} f(x) \pm \mathop {\lim }\limits_{x \to {x_0}} g(x)} \right] \;
\]
These were two basic and important properties of limit now there are another 2 for multiplication and division so try to recall them and then give an example of limit question where you use these properties to find the answer.
Complete step by step solution:
In the given question first we have to describe the properties of limits and then we have to give an example of how we will use these properties to evaluate a limit.
Now let \[{x_0}\] belong to real numbers set and suppose that \[\mathop {\lim }\limits_{x \to {x_0}} f(x)\] and \[\mathop {\lim }\limits_{x \to {x_0}} f(x)\] exist and are finite. If \[c \in \] real number set is a constant,
These are some important properties of limit:
\[
1)\mathop {\lim }\limits_{x \to {x_0}} [cf(x)] = c\left[ {\mathop {\lim }\limits_{x \to {x_0}} f(x)} \right] \\
2)\mathop {\lim }\limits_{x \to {x_0}} [f(x) \pm g(x)] = \left[ {\mathop {\lim }\limits_{x \to {x_0}} f(x) \pm \mathop {\lim }\limits_{x \to {x_0}} g(x)} \right] \\
3)\mathop {\lim }\limits_{x \to {x_0}} [f(x) \times g(x)] = \left[ {\mathop {\lim }\limits_{x \to {x_0}} f(x)} \right] \times \left[ {\mathop {\lim }\limits_{x \to {x_0}} g(x)} \right] \;
\]
\[4)\] If \[\mathop {\lim }\limits_{x \to {x_0}} [g(x)] \] is not equal to \[0\] then we can say that,
\[\mathop {\lim }\limits_{x \to {x_0}} \left[ {\dfrac{{f(x)}}{{g(x)}}} \right] = \left[ {\dfrac{{\mathop {\lim }\limits_{x \to {x_0}} f(x)}}{{\mathop {\lim }\limits_{x \to {x_0}} g(x)}}} \right] \]
The above mentioned properties are really helpful in the computation of limits (keep the conditions stated at the beginning in mind): you can work by splitting limits into smaller and simpler parts.
Now we will solve a limit question where we will be using the above given properties:
Let suppose we have given
I.\[\mathop {\lim }\limits_{x \to 0} [5{e^x}] \]
Now here we will use first property of limit i.e.
\[
\mathop {\lim }\limits_{x \to 0} [5{e^x}] \\
\Rightarrow 5\mathop {\lim }\limits_{x \to 0} [{e^x}] = 5 \;
\]
Because \[\mathop {\lim }\limits_{x \to 0} [{e^x}] = 1\] .
Let suppose we have given
II.\[\mathop {\lim }\limits_{x \to 0} [{e^x} + 5x] \]
Now here we will use first as well as second property i.e.
\[
\mathop {\lim }\limits_{x \to 0} [{e^x} + 5x] \\
\Rightarrow \mathop {\lim }\limits_{x \to 0} [{e^x}] + \mathop {\lim }\limits_{x \to 0} [5x] \;
\]
Using second property we get:
\[
1 + 5\mathop {\lim }\limits_{x \to 0} [x] \\
\Rightarrow 1 + 0 = 1 \;
\]
Let suppose we have given
III.\[\mathop {\lim }\limits_{x \to 2} \left[ {\dfrac{{(4{x^2} + x)(3x)}}{{3x}}} \right] \]
Now here we will use all the given properties to solve this limit i.e.
First we will separate the limits i.e.
\[\dfrac{{\mathop {\lim }\limits_{x \to 2} (4{x^2} + x)\mathop {\lim }\limits_{x \to 2} (3x)}}{{\mathop {\lim }\limits_{x \to 2} (3x)}}\]
Now we will evaluate the limit i.e.
\[
\dfrac{{[4{{(2)}^2} + 2] [3(2)] }}{{3(2)}} \\
\Rightarrow \dfrac{{(4 \times 4 + 2)(6)}}{6} \\
\Rightarrow 16 + 2 = 18 \;
\]
So this is how we use the properties of limits to solve any limit.
Note: Now in the given solution we have mentioned the important properties but there many others as well which you should learn in order to solve a limit. Also you should know how to solve a limit of different functions like log, \[{e^x}\] etc. Be careful with the limits while separating the terms to solve the limit.
Now let \[{x_0}\] belong to real numbers set and suppose that \[\mathop {\lim }\limits_{x \to {x_0}} f(x)\] and \[\mathop {\lim }\limits_{x \to {x_0}} f(x)\] exist and are finite. If \[c \in \] real number set is a constant,
These are some important properties of limit:
\[
1)\mathop {\lim }\limits_{x \to {x_0}} [cf(x)] = c\left[ {\mathop {\lim }\limits_{x \to {x_0}} f(x)} \right] \\
2)\mathop {\lim }\limits_{x \to {x_0}} [f(x) \pm g(x)] = \left[ {\mathop {\lim }\limits_{x \to {x_0}} f(x) \pm \mathop {\lim }\limits_{x \to {x_0}} g(x)} \right] \;
\]
These were two basic and important properties of limit now there are another 2 for multiplication and division so try to recall them and then give an example of limit question where you use these properties to find the answer.
Complete step by step solution:
In the given question first we have to describe the properties of limits and then we have to give an example of how we will use these properties to evaluate a limit.
Now let \[{x_0}\] belong to real numbers set and suppose that \[\mathop {\lim }\limits_{x \to {x_0}} f(x)\] and \[\mathop {\lim }\limits_{x \to {x_0}} f(x)\] exist and are finite. If \[c \in \] real number set is a constant,
These are some important properties of limit:
\[
1)\mathop {\lim }\limits_{x \to {x_0}} [cf(x)] = c\left[ {\mathop {\lim }\limits_{x \to {x_0}} f(x)} \right] \\
2)\mathop {\lim }\limits_{x \to {x_0}} [f(x) \pm g(x)] = \left[ {\mathop {\lim }\limits_{x \to {x_0}} f(x) \pm \mathop {\lim }\limits_{x \to {x_0}} g(x)} \right] \\
3)\mathop {\lim }\limits_{x \to {x_0}} [f(x) \times g(x)] = \left[ {\mathop {\lim }\limits_{x \to {x_0}} f(x)} \right] \times \left[ {\mathop {\lim }\limits_{x \to {x_0}} g(x)} \right] \;
\]
\[4)\] If \[\mathop {\lim }\limits_{x \to {x_0}} [g(x)] \] is not equal to \[0\] then we can say that,
\[\mathop {\lim }\limits_{x \to {x_0}} \left[ {\dfrac{{f(x)}}{{g(x)}}} \right] = \left[ {\dfrac{{\mathop {\lim }\limits_{x \to {x_0}} f(x)}}{{\mathop {\lim }\limits_{x \to {x_0}} g(x)}}} \right] \]
The above mentioned properties are really helpful in the computation of limits (keep the conditions stated at the beginning in mind): you can work by splitting limits into smaller and simpler parts.
Now we will solve a limit question where we will be using the above given properties:
Let suppose we have given
I.\[\mathop {\lim }\limits_{x \to 0} [5{e^x}] \]
Now here we will use first property of limit i.e.
\[
\mathop {\lim }\limits_{x \to 0} [5{e^x}] \\
\Rightarrow 5\mathop {\lim }\limits_{x \to 0} [{e^x}] = 5 \;
\]
Because \[\mathop {\lim }\limits_{x \to 0} [{e^x}] = 1\] .
Let suppose we have given
II.\[\mathop {\lim }\limits_{x \to 0} [{e^x} + 5x] \]
Now here we will use first as well as second property i.e.
\[
\mathop {\lim }\limits_{x \to 0} [{e^x} + 5x] \\
\Rightarrow \mathop {\lim }\limits_{x \to 0} [{e^x}] + \mathop {\lim }\limits_{x \to 0} [5x] \;
\]
Using second property we get:
\[
1 + 5\mathop {\lim }\limits_{x \to 0} [x] \\
\Rightarrow 1 + 0 = 1 \;
\]
Let suppose we have given
III.\[\mathop {\lim }\limits_{x \to 2} \left[ {\dfrac{{(4{x^2} + x)(3x)}}{{3x}}} \right] \]
Now here we will use all the given properties to solve this limit i.e.
First we will separate the limits i.e.
\[\dfrac{{\mathop {\lim }\limits_{x \to 2} (4{x^2} + x)\mathop {\lim }\limits_{x \to 2} (3x)}}{{\mathop {\lim }\limits_{x \to 2} (3x)}}\]
Now we will evaluate the limit i.e.
\[
\dfrac{{[4{{(2)}^2} + 2] [3(2)] }}{{3(2)}} \\
\Rightarrow \dfrac{{(4 \times 4 + 2)(6)}}{6} \\
\Rightarrow 16 + 2 = 18 \;
\]
So this is how we use the properties of limits to solve any limit.
Note: Now in the given solution we have mentioned the important properties but there many others as well which you should learn in order to solve a limit. Also you should know how to solve a limit of different functions like log, \[{e^x}\] etc. Be careful with the limits while separating the terms to solve the limit.
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