Question

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Answer 1

$\Huge\text{\underline{\underline{Explanation}}}$

$\Large\text{[Question 3.]}$

We are solving question number 3.

Let's solve for the first limiting value. In this question, we see an indeterminate limiting value, $\dfrac{0}{0}$.

$\text{ \cdots\longrightarrow\displaystyle\lim_{y\to0}\dfrac{5y^{3}+8y^{2}}{3y^{4}-16y^{2}}=\dfrac{0}{0} }$

So, it is just discontinuous at $y=0$.

As y gets closer to 0, this expression gets closer to $-\dfrac{1}{2}$. Let's see why.

$\text{ \cdots\longrightarrow\displaystyle\lim_{y\to0}\dfrac{y^{2}(5y+8)}{y^{2}(3y^{2}-16)} }$

$\text{ \cdots\longrightarrow\displaystyle\lim_{y\to0}\dfrac{5y+8}{3y^{2}-16} }$

Now, we can substitute to see what value it approaches. As $y\to0$, the limiting value is $\boxed{-\dfrac{1}{2}}$.

$\Large\text{[Question 4.]}$

We have the same type of question in number 4.

$\text{ \cdots\longrightarrow\displaystyle\lim_{x\to-2}\dfrac{-2x-4}{x^{3}+2x^{2}} }$

$\text{ \cdots\longrightarrow\displaystyle\lim_{x\to-2}\dfrac{-2(x+2)}{x^{2}(x+2)} }$

$\text{ \cdots\longrightarrow\displaystyle\lim_{x\to-2}-\dfrac{2}{x^{2}} }$

Now, we can substitute to see what value it approaches. As $x\to-2$, the limiting value is $\boxed{-\dfrac{1}{2}}$.

$\Huge\text{\underline{\underline{Learn more}}}$

$\large\text{Limits involving positive \& negative infinity}$

Let's find more solutions of different limits.

If $x\to\infty$, find the highest degree and divide. Because we know that $\displaystyle\lim_{x\to\infty}\dfrac{1}{x}=0$ we can evaluate the limit.

E.g -

$\text{ \cdots\longrightarrow\displaystyle\lim_{x\to\infty}\dfrac{7x^{2}+8}{3x^{2}+5x+2} }$

$\text{ \cdots\longrightarrow\displaystyle\lim_{x\to\infty}\dfrac{\dfrac{1}{x^{2}}(7x^{2}+8)}{\dfrac{1}{x^{2}}(3x^{2}+5x+2)} }$

$\text{ \cdots\longrightarrow\displaystyle\lim_{x\to\infty}\dfrac{7+\dfrac{8}{x^{2}}}{3+\dfrac{5}{x}+\dfrac{2}{x^{2}}} }$

$\cdots\longrightarrow\displaystyle\lim_{x\to\infty}\dfrac{7+0}{3+0+0}$

$\large\text{ \cdots\longrightarrow\boxed{\begin{aligned}\dfrac{7}{3}\end{aligned}} }$

If $x\to-\infty$, consider the substitution of $t=-x$. This type of question usually involves square roots.

E.g -

$\text{ \cdots\longrightarrow\displaystyle\lim_{x\to-\infty}\dfrac{\sqrt{x^{2}+1}}{\sqrt{4x^{2}+x+1}-x} }$

As $x\to-\infty$, $t\to\infty$.

$\text{ \cdots\longrightarrow\displaystyle\lim_{t\to\infty}\dfrac{\sqrt{t^{2}+1}}{\sqrt{4t^{2}-t+1}+t} }$

Notice the sign changed. Now, we can deal with this limit by dividing by $t$.