# How do you evaluate sin(x-3)/(x^2+4x-21) as x approaches 3?

Mar 23, 2018

$\frac{1}{10}$

#### Explanation:

${\lim}_{x \rightarrow 3} S \in \frac{x - 3}{{x}^{2} + 4 x - 21}$

Using L-Hospital rule,i.e differentiate numerator and denominator separately without using the quotient rule,
we get,

${\lim}_{x \rightarrow 3} C o s \frac{x - 3}{2 x + 4} \cdot \frac{d}{\mathrm{dx}} \left(x - 3\right)$

${\lim}_{x \rightarrow 3} C o s \frac{x - 3}{2 x + 4} \cdot \left(1 - 0\right)$

Putting x=3,

Cos(3-3)/{2(3)+4

=$C o s \frac{0}{10}$
=$\frac{1}{10}$

Mar 23, 2018

The limit is equal to $\frac{1}{10}$.

#### Explanation:

To start, let's just try plugging $x = 3$ into the equation and see what happens:

$\textcolor{w h i t e}{\implies} \sin \frac{x - 3}{{x}^{2} + 4 x - 21}$

$\implies \sin \frac{3 - 3}{{3}^{2} + 4 \left(3\right) - 21}$

$= \sin \frac{0}{9 + 12 - 21}$

$= \frac{0}{0}$

The indeterminate $\frac{0}{0}$. In these types of cases, we can apply l'Hopital's rule:

If ${\lim}_{x \to a} f \frac{x}{g} \left(x\right) = \frac{0}{0}$ or ${\lim}_{x \to a} f \frac{x}{g} \left(x\right) = \frac{\pm \infty}{\pm \infty}$, then

${\lim}_{x \to a} f \frac{x}{g} \left(x\right) = {\lim}_{x \to a} \frac{f ' \left(x\right)}{g ' \left(x\right)}$

In other words, take the derivative of the top and bottom of the fraction, then plug in the value. Let's do that:

$\textcolor{w h i t e}{=} {\lim}_{x \to 3} \sin \frac{x - 3}{{x}^{2} + 4 x - 21}$

$= {\lim}_{x \to 3} \frac{\frac{d}{\mathrm{dx}} \left(\sin \left(x - 3\right)\right)}{\frac{d}{\mathrm{dx}} \left({x}^{2} + 4 x - 21\right)}$

$= {\lim}_{x \to 3} \cos \frac{x - 3}{\frac{d}{\mathrm{dx}} \left({x}^{2} + 4 x - 21\right)}$

$= {\lim}_{x \to 3} \cos \frac{x - 3}{\frac{d}{\mathrm{dx}} \left({x}^{2}\right) + \frac{d}{\mathrm{dx}} \left(4 x\right) - \frac{d}{\mathrm{dx}} \left(21\right)}$

$= {\lim}_{x \to 3} \cos \frac{x - 3}{2 x + 4 - 0}$

$= {\lim}_{x \to 3} \cos \frac{x - 3}{2 x + 4}$

$= \cos \frac{3 - 3}{2 \left(3\right) + 4}$

$= \cos \frac{0}{2 \left(3\right) + 4}$

$= \cos \frac{0}{6 + 4}$

$= \cos \frac{0}{10}$

$= \frac{1}{10}$

That's the limit. We can observe this from the graph of the function:

graph{sin(x-3)/(x^2+4x-21) [-1, 7, -0.02, 0.2]}

Hope this helped!