\(\displaystyle \begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-20sin(7\cdot \pi x)}{ln(x^{8})^{3}}\rightarrow\frac{0}{0}\\&\frac{-140\cdot \pi cos(7\cdot \pi x)}{\frac{(24ln(x^{8})^{2})}{x}}\rightarrow\frac{140\cdot \pi}{0}=\infty\\&lim_{x\rightarrow1+}\frac{-20sin(7\cdot \pi x)}{ln(x^{8})^{3}}=\infty\end{align*}\)

\(\displaystyle \begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{10x}{19ln((x + 1)^{8})^{2}}\rightarrow\frac{0}{0}\\&\frac{10}{\frac{(304ln((x + 1)^{8}))}{(x + 1)}}\rightarrow\frac{10}{0}=\infty\\&lim_{x\rightarrow0+}\frac{10x}{19ln((x + 1)^{8})^{2}}=\infty\end{align*}\)

This question is asking you to recall the limit definition of a derivative, which states:

\(\displaystyle \lim_{h\rightarrow 0} \frac{f(x+h) - f(x)}{h}=f'(x).\)

Therefore, if we can define our function for our problem, we can simply evaluate this limit by taking its derivative.

\(\displaystyle \lim_{h\rightarrow 0} \frac{(x+h)^2 - x^2}{h}.\)  In these types of questions, you look at the second term in the numerator.  That is the negative of the function.

Since \(\displaystyle f(x)=x^2\), \(\displaystyle f'(x)=2x.\)

The first step is to always plug in the value of the limit. Doing so we get 

\(\displaystyle \frac{2(0)^2)}{5}=0\)

Since this is a valid answer, no further work is required. 

The first step is to always plug in the value of the limit. Doing so we get 

\(\displaystyle \frac{5(0)+10}{0+1}=\frac{10}{1}=10\)

Since this is a valid answer, no further work is required. 

The first step is to always plug in the value of the limit. Doing so we get 

\(\displaystyle \frac{3(2)^2+2(2)}{2-1}=\frac{(3)(4)+(2)(2)}{1}=16\)

Since this is a valid answer, no further work is required. 

The first step is to always plug in the value of the limit. Doing so we get 

\(\displaystyle \frac{(3^2-3-6)}{3^2-9}=\frac{0}{0}\)

This is indeterminate. We need to verify it is indeterminate and therefore need take another approach. Let's take a look at the original equation again. 

\(\displaystyle \frac{x^2-x-6}{x^2-9}\)

Taking what we learned in algebra, we know that both the numerator and denominator of the equation are factorable. The equation becomes

\(\displaystyle \frac{(x-3)(x+2)}{(x-3)(x+3)}\)

The \(\displaystyle (x-3)\) from the top and bottom cancels, which then simplifies to 

\(\displaystyle \frac{(x+2)}{(x+3)}\)

Now we evaluate the limit once again using the simplified equation and we get

\(\displaystyle \frac{(3+2)}{(3+3)}=\frac{5}{6}\)

This is a valid answer. 

The first step is to always plug in the value of the limit. Doing so we get 

\(\displaystyle \frac{(4(0)^2-2(0))}{0}=\frac{0}{0}\)

This is indeterminate. We need to verify it is indeterminate and therefore need take another approach. Let's take a look at the original equation again. 

\(\displaystyle \frac{4x^2-2x}{x}\)

Taking what we learned in algebra, we know that the numerator is factorable. Taking out an \(\displaystyle x\) yields

\(\displaystyle \frac{(x)(4x-2)}{(x)}\)

The \(\displaystyle (x)\) from the top and bottom cancels, which then simplifies to 

\(\displaystyle \frac{(4x-2)}{(1)}\)

Now we evaluate the limit once again using the simplified equation and we get

\(\displaystyle 4(0)-2=-2\)

This is a valid answer

\(\displaystyle \begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{41\cdot (x - 1)^{2}}{46ln(x^{8})}\rightarrow\frac{0}{0}\\&\frac{82x - 82}{\frac{368}{x}}\rightarrow\frac{0}{368}=0\\&lim_{x\rightarrow1-}\frac{41\cdot (x - 1)^{2}}{46ln(x^{8})}=0\end{align*}\)

\(\displaystyle \begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{49ln(x^{8})}{48\cdot (x - 1)^{3}}\rightarrow\frac{0}{0}\\&\frac{\frac{392}{x}}{144\cdot (x - 1)^{2}}\rightarrow\frac{392}{0}=\infty\\&lim_{x\rightarrow1-}\frac{49ln(x^{8})}{48\cdot (x - 1)^{3}}=\infty\end{align*}\)