36eb1a…: “As opposed to the windy trapezoid”

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1 : Can this limit be directly computed by limit laws? $\lim _{x\to 0} xe^{1/x}$

yes no

If we are trying to use limit laws to compute this limit, we would first have to use the Product Law to say that $\lim _{x\to 0}xe^{1/x} = \lim _{x\to 0} x \cdot \lim _{x\to 0} e^{1/x}.$ We are only allowed to use this law if both limits exist, so we must check this first. We know from continuity that $\lim _{x\to 0}x = 0.$ However, we also know that $e^{1/x}$ is badly behaved as $x$ approaches $0$ ; going to infinity on one side and zero on the other. Thus the limit $\lim _{x\to 0} e^{1/x}$ does not exist. Therefore, we cannot use the Product Law.

2 : Can this limit be directly computed by limit laws? $\lim _{x\to 0} \frac {2x^3-1}{e^x-1}$

yes no

If we are trying to use limit laws to compute this limit, we would like to use the Quotient Law to say that $\lim _{x\to 0} \frac {2x^3-1}{e^x-1} = \frac {\lim _{x\to 0} 2x^3-1}{\lim _{x\to 0} e^x-1}.$ We are only allowed to use this law if both limits exist and the denominator is not $0$ . We suspect that the limit in the denominator might equal $0$ , so we check this limit. $\begin{align*} \lim_{x\to 0} e^x-1 & = e^{\lim_{x\to 0}x}-1 \\ & = e^0-1 \\ & = 1-1 \\ & = 0. \end{align*}$

This means that the denominator is zero and hence we cannot use the Quotient Law.

3 : Can this limit be directly computed by limit laws? $\lim _{x\to 0} \frac {x+1}{5^{e^x-1}}$

yes no

3.1 : Compute: $\lim _{x\to 0} \frac {x+1}{5^{e^x-1}} = \answer {1}$

If we are trying to use Limit Laws to compute this limit, we would now have to use the Quotient Law to say that $\lim _{x\to 0} \frac {x+1}{5^{e^x-1}} = \frac { \lim _{x\to 0}x+1}{\lim _{x\to 0}5^{e^x-1}}.$ We are only allowed to use this law if both limits exist and the denominator is not $0$ . Let’s check the denominator and numerator separately. First we’ll compute the limit of the denominator: $\begin{align*} \lim_{x\to 0}5^{e^x-1} & = 5^{\lim_{x\to 0}e^x-1}\\ & = 5^{e^{\left(\lim_{x\to 0}x\right)}-1}\\ & = 5^{e^0-1}\\ & = 5^{1-1}\\ & = 5^{0}\\ & = 1 \end{align*}$

Therefore, the limit in the denominator exists and does not equal $0$ . We can use the Quotient Law, so we will compute the limit of the numerator: $\lim _{x\to 0} x + 1 = 1$ Hence $\frac { \lim _{x\to 0}x + 1}{\lim _{x\to 0}5^{e^x-1}} = \frac {1}{1}=1$

4 : Can this limit be directly computed by limit laws? $\lim _{x\to 1}{(x-1)\cdot \ln (x\ln (x))}$

yes no

If we are trying to use limit laws to compute this limit, we would have to use the Product Law to say that $\lim _{x\to 1}{(x-1)\cdot \ln (x\ln (x))}= \lim _{x\to 1}{(x-1)\cdot \lim _{x\to 1}\ln (x\ln (x))}.$ We are only allowed to use this law if both limits exist. Let’s check each limit separately. $\begin{align*} \lim_{x\to 1} (x-1) &= \lim_{x\to 1} (x)-\lim_{x\to 1}(1)\\ &=1-1\\ &=0. \end{align*}$

If we are trying to use limit laws to compute this limit, we would now want to use the Composition Law to say that $\lim \limits _{x\rightarrow 1}\ln (x\ln (x)) = \ln \left (\lim _{x\to 1}(x\ln (x))\right ).$ But we are only allowed to use this law if $\ln (x)$ is continuous at $\lim _{x\to 1}x\ln (x)$ . So we need to compute this limit to determine if $\ln (x)$ is continuous at that value. We can do this using the Product Law.

$\begin{align*} \lim_{x\to 1}x\ln(x) & = \lim_{x\to 1}x\cdot \lim\limits_{x\rightarrow 1} \ln(x)\\ & = 1\cdot 0\\ & = 0 \end{align*}$

So, since $\lim _{x\to 1}x\ln (x)$ is $0$ , and $\ln (x)$ doesn’t even exist (let alone be continuous) at $0$ , we cannot apply the Composition Law.

5 : Can this limit be directly computed by limit laws? $\lim _{x\to 0} x\ln x$

yes no

If we are trying to use limit laws to compute this limit, we would have to use the Product Law to say that $\lim _{x\to 0} x\ln x =\lim _{x\to 0} x \cdot \lim _{x\to 0}\ln x.$ We are only allowed to use this law if both limits exist. We know $\lim _{x\to 0} x = 0$ , but what about $\lim _{x\to 0}\ln x$ ? We do not know how to find $\lim _{x\to 0}\ln x$ using limit laws because $0$ is not in the domain of $\ln x$ .

6 : Can this limit be directly computed by limit laws? $\lim _{x\to 0}\frac {2^x-1}{3^{x-1}}$

yes no

6.1 : Compute: $\lim _{x\to 0}\frac {2^x-1}{3^{x-1}}\begin {prompt} =\answer {0}\end {prompt}$

If we are trying to use limit laws to compute this limit, we would have to use the Quotient Law to say that $\lim _{x\to 0}\frac {2^x-1}{3^{x-1}} = \frac {\lim _{x\to 0}(2^x-1)}{\lim _{x\to 0}(3^{x-1})}.$ We are only allowed to use this law if both limits exist and the denominator does not equal $0$ . Let’s check each limit separately, starting with the denominator $\begin{align*} \lim_{x\to 0}(3^{x-1}) &=3^{\lim_{x\to0}(x-1)}\\ &=3^{-1}\\ &=\frac{1}{3} \end{align*}$

On the other hand the limit in the numerator is

$\begin{align*} \lim_{x\to 0}(2^x-1) &=\lim_{x\to0}(2^x)-\lim_{x\to0}(1)\\ &=1-1\\ &=0 \end{align*}$

The limits in both the numerator and denominator exist and the limit in the denominator does not equal $0$ , so we can use the Quotient Law. We find: $\frac {\lim _{x\to 0}(2^x-1)}{\lim _{x\to 0}(3^{x-1})} =\frac {0}{\frac {1}{3}}=0.$

7 : Can this limit be directly computed by limit laws? $\lim _{x\to 0}(1+x)^{1/x}$

yes no

We do not have any limit laws for functions of the form $f(x)^{g(x)}$ , so we cannot compute this limit.