
How do you prove the limit of $14-5x=4$ as $x\to 2$ using the precise definition of a limit?
Answer
558.9k+ views
Hint: We first try to find the function and approaching value of the variable x. Then we find the definition of limit and how it applies for the function to find the limit value. The limit only exists when the left-hand and right-hand each limit gives equal value. The mathematical form being $\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$.
Complete step by step answer:
We need to prove that limit of $14-5x=4$ as $x\to 2$. Therefore,
$\underset{x\to 2}{\mathop{\lim }}\,\left( 14-5x \right)=4$.
Let’s assume the function as $f\left( x \right)=\left( 14-5x \right)$.
For our given limit the value of variable x tends to the point 2. This means the value can be approaching from the both sides of the point of 2. We can break it into three parts of ${{2}^{+}},2,{{2}^{-}}$.
${{2}^{+}}$ represents that the value is approaching from the right-side or greater side of the point and ${{2}^{-}}$ represents that the value is approaching from the left-side or lesser side of the point. There is also the fixed point of 2.
Now the limit value will exist only when
$\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)=f\left( 2 \right)$.
We need to find the values of the given function for the approaching value of x.
If they all are equal then we can say the limit value is equal to that number.
$ \Rightarrow \underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right) $
$ \Rightarrow {{\left[ 14-5x \right]}_{x=2}}=14-5\times 2=14-10=4$
$ \Rightarrow \underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)$
$ \Rightarrow {{\left[ 14-5x \right]}_{x=2}}=14-5\times 2=14-10=4$
$ \Rightarrow f\left( 2 \right)=14-5\times 2=14-10=4$
All the values are equal to 4 which proves the limit of $14-5x=4$ as $x\to 2$.
Note: The precise definition of a limit is something we use as a proof for the existence of a limit. When we’re evaluating a limit, we’re looking at the function as it approaches a specific point. we approach a particular value of x, the function itself gets closer and closer to a particular value.
Complete step by step answer:
We need to prove that limit of $14-5x=4$ as $x\to 2$. Therefore,
$\underset{x\to 2}{\mathop{\lim }}\,\left( 14-5x \right)=4$.
Let’s assume the function as $f\left( x \right)=\left( 14-5x \right)$.
For our given limit the value of variable x tends to the point 2. This means the value can be approaching from the both sides of the point of 2. We can break it into three parts of ${{2}^{+}},2,{{2}^{-}}$.
${{2}^{+}}$ represents that the value is approaching from the right-side or greater side of the point and ${{2}^{-}}$ represents that the value is approaching from the left-side or lesser side of the point. There is also the fixed point of 2.
Now the limit value will exist only when
$\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)=f\left( 2 \right)$.
We need to find the values of the given function for the approaching value of x.
If they all are equal then we can say the limit value is equal to that number.
$ \Rightarrow \underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right) $
$ \Rightarrow {{\left[ 14-5x \right]}_{x=2}}=14-5\times 2=14-10=4$
$ \Rightarrow \underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)$
$ \Rightarrow {{\left[ 14-5x \right]}_{x=2}}=14-5\times 2=14-10=4$
$ \Rightarrow f\left( 2 \right)=14-5\times 2=14-10=4$
All the values are equal to 4 which proves the limit of $14-5x=4$ as $x\to 2$.
Note: The precise definition of a limit is something we use as a proof for the existence of a limit. When we’re evaluating a limit, we’re looking at the function as it approaches a specific point. we approach a particular value of x, the function itself gets closer and closer to a particular value.
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