# Evaluate the limit. \lim_{ x \rightarrow 2}\arctan((2x^{2}-8)/(3x^{2}-6x))

## Question:

Evaluate the limit.

{eq}\displaystyle \lim_{ x \rightarrow 2}\arctan((2x^{2}-8)/(3x^{2}-6x)) {/eq}

## Limit of a Function:

Let {eq}f {/eq} be a given function. Then a real number {eq}l {/eq} is said to be the limit of the function {eq}f {/eq} at a point {eq}a {/eq} in the domain of {eq}f {/eq} if for each {eq}\epsilon>0 {/eq} there exists a {eq}\delta >0 {/eq} such that {eq}\left| f(x) - l \right| < \epsilon {/eq} for all {eq}x {/eq} satisfying {eq}\left| {x - a} \right| < \delta {/eq} and donted by {eq}\mathop {\lim }\limits_{x \to a} f(x) {/eq}. This is also known as the {eq}\epsilon - \delta {/eq} definition of a limit.

#### Applications of the L'Hospital's Rule:

L'Hospital rule mainly used to determine the limit of some special types of indeterminate forms. This can be stated as follows:

Let {eq}{f_1}\left( x \right),\,\,{f_2}\left( x \right) {/eq} be two differentiable functions of {eq}x {/eq}. Suppose we have one of following cases at {eq}x=c {/eq}, {eq}\displaystyle \frac{{{f_1}\left( c \right)}}{{{f_2}\left( c \right)}} = \frac{0}{0} {/eq} or

{eq}\displaystyle \frac{{{f_1}\left( c \right)}}{{{f_2}\left( c \right)}} = \frac{{ \pm \infty }}{{ \pm \infty }} {/eq}

Then the limiting behaviour at {eq}x=c {/eq} of {eq}\displaystyle \frac{f_1(x)}{f_2(x)} {/eq} is same as {eq}\displaystyle \frac{f'_1(x)}{f'_2(x)} {/eq}. That is according to the L'Hospital Rule, {eq}\displaystyle \mathop {\lim }\limits_{x \to c} \frac{{{f_1}\left( x \right)}}{{{f_2}\left( x \right)}} = \mathop {\lim }\limits_{x \to c} \frac{{{f_1}'\left( x \right)}}{{{f_2}'\left( x \right)}} {/eq}.

Here we have to find the value of limit {eq}\displaystyle \mathop {\lim }\limits_{x \to 2} \arctan \frac{{2{x^2} - 8}}{{3{x^2} - 6x}} {/eq}

Now let us first evaluate the following limit: {eq}\displaystyle \mathop {\lim }\limits_{x \to 2} \frac{{2{x^2} - 8}}{{3{x^2} - 6x}} {/eq}

Here we can see that {eq}\displaystyle x \to 2 \Rightarrow 2{x^2} - 8 \to 0\,\,\,\,{\text{and }}\,3{x^2} - 6x \to 0 {/eq}, so the given limit is in {eq}\displaystyle {\frac{0}{0}\,\,\,{\text{form}}} {/eq}, using the L'Hospital Rule, we get

{eq}\displaystyle \mathop {\lim }\limits_{x \to 2} \frac{{2{x^2} - 8}}{{3{x^2} - 6x}} = \mathop {\lim }\limits_{x \to 2} \frac{{4x}}{{6x - 6}} = \frac{8}{{12 - 6}} = \frac{8}{6} = \frac{4}{3}. {/eq}

Thus we have {eq}\displaystyle \mathop {\lim }\limits_{x \to 2} \arctan \frac{{2{x^2} - 8}}{{3{x^2} - 6x}} = \arctan \left( {\frac{4}{3}} \right). {/eq}

Therefore we can say that the value of the limit is {eq}\displaystyle \arctan \left( {\frac{4}{3}} \right). {/eq}