$\text { Find the limit using } \lim _ { x = 0 } \frac { \sin x } { x } = 1 \text {. }$
-$$\lim _ { x \rightarrow 1 ^ { + } } \left( \frac { x } { x + 9 } \right) \left( \frac { - 3 x + 8 } { x ^ { 2 } + 9 x } \right)$$
A) Does not exist
B) $\frac { 11 } { 82 }$
C) $\frac { 11 } { 64 }$
D) $\frac { 1 } { 11 }$

Question

Quizplus · Accepted Answer

First, notice that the denominator of the second fraction does not equal 0, whereas the denominator of the first fraction equals 10. Thus, the entire expression is defined on an open interval $(1-\delta,1+\delta)$ for some $\delta > 0$.

Next, we can simplify the expression using algebraic manipulation:

\begin{align*}
\lim_{x	o 1^+} \left(\frac{x}{x+9}ight)\left(\frac{-3x+8}{x^2+9x}ight) &= \lim_{x	o 1^+} \left(\frac{x(-3x+8)}{(x+9)(x^2+9x)}ight)\
&= \lim_{x	o 1^+} \left(\frac{-3x^2+8x}{x(x^2+9x+9\cdot 1)}ight)\
&= \lim_{x	o 1^+} \left(\frac{-3x+8}{x^2+9x+9}ight)
\end{align*}

Substituting $x = 1$ into the expression, we find that the limit equals $\frac{5}{19}$. However, this is not the desired limit. Instead, we need to evaluate the limit as $x$ approaches 1 from the right.

One way to evaluate this limit is to factorize the denominator and apply partial fraction decomposition:

$$x^2+9x+9 = (x+9)\left(x+\frac{1}{2}ight)^2-\frac{99}{4}$$

Thus, we can write

$$\frac{-3x+8}{x^2+9x+9} = \frac{A}{x+9}+\frac{Bx+C}{x^2+9x+9}$$

for some constants $A$, $B$, and $C$ that we can solve for by equating numerators:

$$-3x+8 = A(x^2+9x+9)+(Bx+C)(x+9)$$

Substituting $x=1$, we obtain $A = -\frac{11}{64}$. The other constants are not needed for our purposes.

Thus, we can rewrite the limit as

\begin{align*}
\lim_{x	o 1^+} \left(\frac{-3x+8}{x^2+9x+9}ight) &= \lim_{x	o 1^+} \left(\frac{-11/(64)}{x+9}ight)+\lim_{x	o 1^+} \left(\frac{Bx+C}{x^2+9x+9}ight)\
&= \frac{-11}{64}\lim_{x	o 1^+} \left(\frac{1}{x+9}ight)+\lim_{x	o 1^+} \left(\frac{Bx+C}{x^2+9x+9}ight)\
&= \frac{-11}{64}\cdot \frac{1}{10}+0\
&= \boxed{\frac{11}{640}}
\end{align*}

Find the limit using lim⁡x=0sin⁡xx=1. \text { Find the limit using } \lim _ { x = 0 } \frac { \sin x } { x } = 1 \text {. } Find the limit using limx=0​xsinx​=1.

$\text { Find the limit using } \lim _ { x = 0 } \frac { \sin x } { x } = 1 \text {. }$