Question:
$\lim_{x \to {1}} [\frac {x+2} {x^2 -5x +4} + \frac {x-4} {3(x^2-3x+2)}]$
$\lim_{x \to {1}} [\frac {x+2} {x^2 -5x +4} + \frac {x-4} {3(x^2-3x+2)}]$
Updated On: Jul 6, 2022
- 0
- $\frac {1} {6}$
- $\frac {1} {3}$
- 1
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The Correct Option is A
Solution and Explanation
$\lim_{x\to1}\left[\frac{x+2}{x^{2} -5x+4}+\frac{x-4}{3\left(x^{4}-3x +2\right)}\right]$
= $\lim_{x\to1} \left[\frac{x+2}{\left(x-1\right)\left(x-2\right)}+\frac{x-4}{3\left(x-1\right)\left(x-2\right)}\right]$
= $\lim_{x\to1} \left[\frac{3\left(x^{2}-4\right)+\left(x-4\right)^{2}}{3\left(x-1\right)\left(x-2\right)\left(x-4\right)}\right]$
= $\lim_{x\to1}\left[\frac{4x^{2 }-8x +4}{3\left(x-1\right)\left(x-2\right)\left(x-4\right)}\right]$
= $\frac{4}{3} \lim_{x\to1} \frac{\left(x-1\right)^{2}}{3\left(x-1\right)\left(x-2\right)\left(x-4\right)}$
= $ \frac{4}{3} \lim_{x\to1} \frac{x-1}{\left(x-2\right)\left(x-4\right)}$
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Concepts Used:
Limits
A function's limit is a number that a function reaches when its independent variable comes to a certain value. The value (say a) to which the function f(x) approaches casually as the independent variable x approaches casually a given value "A" denoted as f(x) = A.
If limx→a- f(x) is the expected value of f when x = a, given the values of ‘f’ near x to the left of ‘a’. This value is also called the left-hand limit of ‘f’ at a.
If limx→a+ f(x) is the expected value of f when x = a, given the values of ‘f’ near x to the right of ‘a’. This value is also called the right-hand limit of f(x) at a.
If the right-hand and left-hand limits concur, then it is referred to as a common value as the limit of f(x) at x = a and denote it by lim x→a f(x).