Question

Indicate for which value of function x is not constant:
$\purple{ \mathtt{\boxed{ \mathtt{f ( x ) = \frac { 3 x + 7 } { x ^ { 2 } - 5 x + 6 }}}}}$
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Answer 1

$\Huge\red{\underline{\red{\bold{Limit\ Basics}}}}$

$\Large{\text{\underline{1. LHL and RHL}}}$

Firstly, LHL and RHL are approaching values towards a point. The values of both are meant in the graphical approach. For example, refer to the attachment.

• $\displaystyle\lim_{x\to a-}f(x)\large\text{ (LHL; Limiting value approaching from the left.)}$
• $\displaystyle\lim_{x\to a+}f(x)\large\text{ (RHL; Limiting value approaching from the right.)}$

If two values are the same, we refer to this value as $\displaystyle\lim_{x\to a}f(x)$, the limit of $f(x)$ as $x$ goes to $a$.

$\Large{\text{\underline{2. Continuity}}}$

That a function is continuous is meant by the points of the points are smoothly joined. We define continuity by the LHL(left-hand limit) and RHL(right-hand limit) and the functional value.

$\Huge\red{\underline{\red{\bold{Solution}}}}$

$f(x)$ is a rational function if we classify it. The domain of the graph does not include the zeros of the denominator, as numbers in the form of $\dfrac{k}{0}$ are undefined.

$\large\red{\boxed{\red{\bold{f(x)=\dfrac{3x+7}{(x-2)(x-3)}}}}}$

So, the graph is discontinuous at $x=2,3$. However, we have learned about continuousity so let's make use of it.

$\displaystyle\large\red{\boxed{\red{\bold{\lim_{x\to2-}f(x)=\lim_{x\to2-}\dfrac{11}{(-1)(x-2)}=\infty}}}}$

$\displaystyle\large\red{\boxed{\red{\bold{\lim_{x\to2+}f(x)=\lim_{x\to2+}\dfrac{11}{(-1)(x-2)}=-\infty}}}}$

The limiting values are different. So, $\displaystyle\lim_{x\to2}f(x)$ does not exist and the function is discontinuous at $x=-2$. (Moreover, they do not converge. Infinity is not a value, but a situation where the value keeps increasing as it approaches.)

$\displaystyle\large\red{\boxed{\red{\bold{\lim_{x\to3-}f(x)=\lim_{x\to3-}\dfrac{11}{(1)(x-3)}=-\infty}}}}$

$\displaystyle\large\red{\boxed{\red{\bold{\lim_{x\to3+}f(x)=\lim_{x\to3+}\dfrac{11}{(1)(x-3)}=\infty}}}}$

And,

$\displaystyle\large\red{\boxed{\red{\bold{\lim_{x\to-\infty}f(x)=\lim_{x\to\infty}\dfrac{\dfrac{-3x}{x^{2}}+\dfrac{7}{x^{2}}}{\dfrac{x^{2}}{x^{2}}-\dfrac{-5x}{x^{2}}+\dfrac{6}{x^{2}}}=0}}}}$

$\displaystyle\large\red{\boxed{\red{\bold{\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\dfrac{\dfrac{3x}{x^{2}}+\dfrac{7}{x^{2}}}{\dfrac{x^{2}}{x^{2}}-\dfrac{5x}{x^{2}}+\dfrac{6}{x^{2}}}=0}}}}$

$\Huge\red{\underline{\red{\bold{Learn\ More}}}}$

We can guess the graph of $f(x)$ also. Using the limiting values, we get the graph in the attachment.