Calculus/Limits/Solutions

Since this is a polynomial, two can simply be plugged in. This results in ${\displaystyle 4(4)-2(3)-1=16-6-1=\mathbf {9} }$

Factor as ${\displaystyle {\frac {x^{2}}{x^{2}}}\cdot {\frac {x+1}{x+2}}}$ . In this form we can see that there is a removable discontinuity at ${\displaystyle x=0}$ and that the limit is ${\displaystyle \mathbf {\frac {1}{2}} }$

${\displaystyle {\sqrt {1-x^{2}}}}$ is defined if ${\displaystyle x^{2}<1}$, so the limit is ${\displaystyle {\sqrt {1-1^{2}}}=\mathbf {0} }$

${\displaystyle {\sqrt {1-x^{2}}}}$ is not defined if ${\displaystyle x^{2}>1}$ , so the limit does not exist.

If approaching from the left, then ${\displaystyle |x|=-x}$. Thus, ${\displaystyle \lim _{x\to 0^{-}}{\frac {|x|}{x}}={\frac {-x}{x}}=\mathbf {-1} }$

${\displaystyle \arcsin(x)=\sin ^{-1}(x)}$ is defined from ${\displaystyle x\in \left[-1,1\right]}$, so the limit is ${\displaystyle \arcsin(1)=\mathbf {\frac {\pi }{2}} }$.

${\displaystyle {\begin{aligned}&\lim _{x\to 4^{-}}{\frac {1}{x-4}}=-\infty \\&\lim _{x\to 4^{+}}{\frac {1}{x-4}}=\infty \end{aligned}}}$
The limit does not exist.

As ${\displaystyle x}$ approaches ${\displaystyle 2}$, the denominator will be a very small positive number, so the whole fraction will be a very large positive number. Thus, the limit is ${\displaystyle \mathbf {\infty } }$.

As ${\displaystyle x}$ approaches ${\displaystyle 3}$, the numerator goes to 5 and the denominator goes to 0. Depending on whether you approach ${\displaystyle 3}$ from the left or the right, the denominator will be either a very small negative number, or a very small positive number. So the limit from the left is ${\displaystyle -\infty }$ and the limit from the right is ${\displaystyle +\infty }$. Thus, the limit does not exist.

The numerator evaluates to ${\displaystyle 0}$ while the denominator evaluates to ${\displaystyle -3}$. Hence, ${\displaystyle \lim _{x\to 0}{\frac {x^{2}}{x^{2}+2x-3}}=\mathbf {0} }$.

Notice that as ${\displaystyle x}$ approaches ${\displaystyle 1}$, the numerator approaches ${\displaystyle 5}$ while the denominator approaches ${\displaystyle 0}$. However, if you approach from below, the denominator is negative, and if you approach from above, the denominator is positive. So the limits from the left and right will be ${\displaystyle -\infty }$ and ${\displaystyle +\infty }$ respectively. Thus, the limit does not exist.

Notice how ${\displaystyle \lim _{x\to -1}\ln \left({\sqrt {1-x^{2}}}\right)=\ln \left(\lim _{x\to -1}{\sqrt {x^{2}-1}}\right)}$.

${\displaystyle \ln \left(\lim _{x\to -1^{+}}{\sqrt {x^{2}-1}}\right)=\ln({\text{DNE}})={\text{DNE}}}$

${\displaystyle \ln \left(\lim _{x\to -1^{-}}{\sqrt {x^{2}-1}}\right)=\ln \left(0^{+}\right)=\infty }$

Thus, the limit does not exist.

Notice that

${\displaystyle \lim _{x\to 1^{+}}\arcsin(x)}$ does not exist since the domain of ${\displaystyle \arcsin(x)}$ is ${\displaystyle x\in \left[-1,1\right]}$

${\displaystyle \lim _{x\to 1^{-}}\arcsin(x)={\frac {\pi }{2}}}$

Thus, the limit does not exist.

This rational function is bottom-heavy, so the limit is ${\displaystyle \mathbf {0} }$.

This rational function has evenly matched powers of ${\displaystyle x}$ in the numerator and denominator, so the limit will be the ratio of the coefficients, i.e. ${\displaystyle \mathbf {\frac {1}{3}} }$.

Balanced powers in the numerator and denominator, so the limit is the ratio of the coefficients, i.e. ${\displaystyle \mathbf {\frac {3}{2}} }$.

This is a top-heavy rational function, where the exponent of the ratio of the leading terms is ${\displaystyle 2}$. Since it is even, the limit will be ${\displaystyle \mathbf {\infty } }$.

Bottom-heavy rational function, so the limit is ${\displaystyle \mathbf {0} }$.

This is a rational function, as can be seen by writing it in the form ${\displaystyle {\frac {6x^{0}}{1x^{0}}}}$. Since the powers of ${\displaystyle x}$ in the numerator and denominator are evenly matched, the limit will be the ratio of the coefficients, i.e. ${\displaystyle \mathbf {6} }$.

Bottom-heavy, so the limit is ${\displaystyle \mathbf {0} }$.

Evenly matched highest powers of ${\displaystyle x}$ in the numerator and denominator, so the limit will be the ratio of the corresponding coefficients, i.e. ${\displaystyle \mathbf {\frac {3}{2}} }$.

Top-heavy rational function, where the exponent of the ratio of the leading terms is ${\displaystyle 1}$, so the limit is ${\displaystyle \mathbf {-\infty } }$.

Bottom-heavy, so the limit is ${\displaystyle \mathbf {0} }$.

${\displaystyle \lim _{x\to \infty }{\frac {1}{x}}=0}$, so ${\displaystyle \lim _{x\to \infty }\sin \left({\frac {1}{x}}\right)=0}$. From this, ${\displaystyle \lim _{x\to \infty }{\frac {\sin \left({\frac {1}{x}}\right)}{x}}={\frac {0}{\infty }}}$. Because ${\displaystyle \lim _{x\to \infty }{\frac {n}{x}}=0}$ for any ${\displaystyle n\in \left(-\infty ,\infty \right)}$, ${\displaystyle \lim _{x\to \infty }{\frac {\sin \left({\frac {1}{x}}\right)}{x}}=\mathbf {0} }$.

Notice that ${\displaystyle \lim _{x\to \infty }\arctan(x)={\frac {\pi }{2}}}$. Because ${\displaystyle \arctan(-x)=-\arctan(x)}$, ${\displaystyle \lim _{x\to \infty }{\frac {\sin(\arctan(x))}{\arctan(-x)}}={\frac {\sin \left({\frac {\pi }{2}}\right)}{-{\frac {\pi }{2}}}}={\frac {1}{-{\frac {\pi }{2}}}}=\mathbf {-{\frac {2}{\pi }}} }$

Since the limits from the left and right don't match, the limit does not exist.

Since the left and right limits match, the overall limit is also ${\displaystyle \mathbf {1} }$.

Since the limits from the right and left match, the overall limit is ${\displaystyle \mathbf {1} }$. Note that in this case, the limit at 2 does not match the function value at 2, so the function is discontinuous at this point, hence the function is nondifferentiable at this point as well.

Notice ${\displaystyle f(x)}$ is continuous from ${\displaystyle x\in \left[1,9\right]}$. Ergo, the intermediate value theorem applies. For all ${\displaystyle y\in \left[2,{\frac {10}{3}}\right]}$, there exists a ${\displaystyle c\in (1,9)}$ so that ${\displaystyle f(c)=3}$.

Notice ${\displaystyle f(x)}$ is continuous from ${\displaystyle x\in \left[\pi ,{\frac {9}{4}}\pi \right]}$. Ergo, the intermediate value theorem applies.

${\displaystyle f(\pi )={\frac {1}{\ln(\pi )}}}$

${\displaystyle f\left({\frac {9}{4}}\pi \right)=0}$

It is known the following is true: ${\displaystyle 1\geq {\frac {\pi }{4}}\geq 0}$. From there, we can directly argue the following:

${\displaystyle {\begin{aligned}{\frac {1}{\ln(\pi )}}\geq 1&\Rightarrow {\frac {1}{\ln(\pi )}}\geq 1\geq {\frac {\pi }{4}}\geq 0\\&\Rightarrow {\frac {1}{\ln(\pi )}}\geq {\frac {\pi }{4}}\geq 0\end{aligned}}}$

By the intermediate value theorem, if ${\displaystyle f(x)}$ is continuous from ${\displaystyle x\in \left[\pi ,{\frac {9}{4}}\pi \right]}$, then there exists an ${\displaystyle x=\mu }$ so that ${\displaystyle f\left({\frac {9}{4}}\pi \right)\leq f(\mu )\leq f(\pi )}$ for ${\displaystyle \mu \in \left[\pi ,{\frac {9}{4}}\pi \right]}$. ${\displaystyle \blacksquare }$

Notice ${\displaystyle f(x)}$ is not continuous for ${\displaystyle x=0}$ since ${\displaystyle \lim _{x\to 0}f(x)}$ is unbounded. Ergo, the intermediate value theorem cannot be used to solve this problem.