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# How do you use the formal definition of a limit to prove lim (${x^2} - x + 1$ )=1 as x approaches 1?

Last updated date: 17th Jun 2024
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Hint: In order to solve the above problem we must understand the limit.
Limit is the value that a function approaches as the input approaches some value. Limits are essential to calculus and mathematical analysis.
Using the method to solve limits, we will solve the given problem.

Let's first discuss more about limits and then we will calculate the value of the given limit function.
For any function, the limit to exist we have to make, the value of the given function at the defined limit equal to the right hand limit and left hand limit.
$\mathop {\lim }\limits_{h \to {0^ - }} f(c - h) = \mathop {\lim f(c + h) = f(c)}\limits_{h \to {0^ + }}$ (left hand limit, right hand limit and the value of the function must be equal for the limit to exist)
First we will check whether the given function is in indeterminate form or not.
$\Rightarrow \mathop {\lim (}\limits_{x \to 1} {x^2} - x + 1)$ the given function does not contain any fraction part so the function is not in determinant
We can easily apply the limit now;
First we will substitute the value of limit in the function,
$\Rightarrow ({1^2} - 1 + 1)$(We have removed the limit and substituted the value of 1)
$\Rightarrow 1$
Now the left hand limit is;
$\mathop { \Rightarrow \lim }\limits_{h \to {0^ - }} ({(c - h)^2} - (c - h) + 1)$
After opening the bracket
$\mathop { \Rightarrow \lim }\limits_{h \to {0^ - }} (({h^2} - h + 1)$
$\Rightarrow 1$
Now the right hand limit;
$\mathop { \Rightarrow \lim }\limits_{h \to {0^ - }} ({(c + h)^2} - (c + h) + 1)$
After opening the bracket
$\mathop { \Rightarrow \lim }\limits_{h \to {0^ - }} ({h^2} - h + 1)$
$\Rightarrow 1$
We have obtained all the three values equal to 1, hence the limit of the function is equal to 1 proved.

Note: When the given function is of in determinant form then the limit of a function could be found out by using the L'HOSPITAL RULE. In order to apply the L'HOSPITAL RULE we have to differentiate both the numerator and denominator and then the value of limit is substituted.