AP Calculus BC - Limits

Evaluate the limit:

$\lim_{x \to 3}\frac{x^2-x-6}{x^2-9}$

$\frac{5}{6}$

$\infty$

Explanation

The limiting situation in this equation would be the denominator. Plug the value that is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will equal zero when x=3; so we try to eliminate the denominator by factoring. When the denominator is no longer zero, we may continue to insert the value of x into the remaining equation.

$\lim_{x \to 3}\frac{x^2-x-6}{x^2-9}=\lim_{x \to 3}\frac{(x-3)(x+2)}{(x-3)(x+3)}=\lim_{x \to 3}\frac{x+2}{x+3}=\frac{3+2}{3+3}=\frac{5}{6}$

Evaluate the limit:

$\lim_{x \to 0}x^2+2x+4$

$\infty$

$-\infty$

Explanation

There is no limiting situation in this equation (like a denominator) so we can just plug in the value that n approaches into the limit and solve:

$\lim_{x \to 0}x^2+2x+4=0^2+2*0+4=4$

$\begin{align*}&\text{Evaluate the following limit:}\&lim_{x\rightarrow0-}\frac{-9tan(18x)}{ln((x + 1)^{4})^{3}}\end{align*}$

$-\infty$

$\infty$

$9\cdot \pi$

$-9\cdot \pi$

Explanation

$\begin{align*}&\text{Examining this problem, we find that both}\&\text{numerator and denominator have zero values at the limit specified.}\&\text{Noting this, L'Hopital's method may be of use:}\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\&\frac{-9tan(18x)}{ln((x + 1)^{4})^{3}}\rightarrow\frac{0}{0}\&\frac{- 162tan(18x)^{2} - 162}{\frac{(12ln((x + 1)^{4})^{2})}{(x + 1)}}\rightarrow\frac{-162}{0}=-\infty\&lim_{x\rightarrow0-}\frac{-9tan(18x)}{ln((x + 1)^{4})^{3}}=-\infty\end{align*}$

Evaluate the following limit:

$\lim_{x\rightarrow 2^{-}}f(x), f(x)=\left{\begin{matrix} \ln(2-x), x\leq 2\ x-3, x> 2 \end{matrix}\right.$

$-\infty$

$\infty$

Explanation

To evaluate the limit, we must determine whether it is right or left sided. The negative sign "exponent" on the 2 indicates that we are approaching from the left, or numbers slightly less than 2. So, we choose the part of the piecewise function that is for x values less than or equal to 2. Now, evaluating the limit, as natural log approaches zero, we get $-\infty$ .

Screen shot 2015 07 22 at 11.03.39 am

Given the above graph of , what is $\lim_{x\rightarrow 0^{+}}f(x)$ ?

$\infty$

$-\infty$

Explanation

Examining the graph, we want to find where the graph tends to as it approaches zero from the right hand side. We can see that there appears to be a vertical asymptote at zero. As the x values approach zero from the right, the function values of the graph tend towards positive infinity.

Therefore, we can observe that $\lim_{x\rightarrow 0^{+}}f(x)=\infty$ as approaches from the right.

Evaluate the limit:

$\lim_{x \to 0}\frac{x^2+2x+1}{x+6}$

$\frac{1}{6}$

Does not exist

$-\infty$

$\infty$

Explanation

The limiting situation in this equation would be the denominator. Plug the value that x is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will not equal zero when x=0; so we proceed to insert the value of x into the entire equation.

$\lim_{x \to 0}\frac{x^2+2x+1}{x+6}=\frac{0^2+2*0+1}{0+6}=\frac{1}{6}$

Evaluate the limit:

$\lim_{n \to \infty}\frac{n^3}{4n^3+n^2+n}$

$\frac{1}{4}$

$\infty$

$-\infty$

Explanation

To evaluate the limit, we must factor out a term consisting of the highest power term divided by itself (which equals one, so we aren't changing the original function):

$\lim_{n \to \infty}(\frac{n^3}{n^3})\frac{1}{4+n^{-1}+n^{-2}}$

The term we factored goes to one, and the two terms with negative exponents in the denominator go to zero (they are each "fractions" with n in their denominator - the terms go to zero as the denominator goes to infinity), so we are left with $\frac{1}{4}$ .

$\begin{align*}&\text{Evaluate the following limit:}\&lim_{x\rightarrow1-}\frac{41\cdot (x - 1)^{2}}{46ln(x^{8})}\end{align*}$

$\infty$

$-\infty$

$\frac{41}{135424}$

$-\frac{41}{135424}$

Explanation

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

Evaluate the limit

$\lim_{n \to 1}\frac{n^2-2n+1}{n^2-1}$

$\infty$

$-\infty$

Explanation

The limiting situation in this equation would be the denominator. Plug the value that n is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will equal zero when n=1; so we try to eliminate the denominator by factoring. When the denominator is no longer zero, we may continue to insert the value of n into the remaining equation.

$\lim_{n \to 1}\frac{n^2-2n+1}{n^2-1}=\lim_{n \to 1}\frac{(n-1)(n-1)}{(n-1)(n+1)}=\lim_{n \to 1}\frac{n-1}{n+1}=\frac{1-1}{1+1}=\frac{0}{2}=0$