$$ \begin{align} \lim_{x\to 0} \frac{sin(x)}{x} \; &\longrightarrow \left( \frac{0}{0} \right) \\[5pt] \lim_{x\to 0} \frac{cos(x)}{1} &= \lim_{x\to 0} \, cos(x) \\[5pt] &= cos(0) = 1 \end{align}$$

$$ \begin{align} \lim_{x\to 0} \frac{sin(x)}{x} \; &\longrightarrow \left( \frac{0}{0} \right) \\[5pt] \lim_{x\to 0} \frac{-sin(x)}{1} &= -\lim_{x\to 0} \, sin(x) \\[5pt] &= sin(0) = 0 \end{align}$$

$$ \begin{align} \lim_{x\to 0} \frac{x - sin(x)}{tan(x)} \; &\longrightarrow \left( \frac{0}{0} \right) \\[5pt] \lim_{x\to 0} \frac{1-cos(x)}{sec^2(x)} &= \lim_{x\to 0} \, \frac{1 - cos(x)}{\left( \frac{1}{cos(x)} \right)^2} \\[5pt] &= \frac{1 - 1}{1} = 0 \end{align}$$

We have to be careful with this function. Take a look at its graph:

Notice that the limits at x = 0 are different from the left and the right. We can work it out using L'Hopital's rule:

$$ \begin{align} \lim_{x\to 0} \frac{ln(cos(x))}{x} \; &\longrightarrow \left( \frac{0}{0} \right) \\[5pt] \lim_{x\to 0} \frac{\frac{1}{cos(x)}(-sin(x))}{1} &= \lim_{x\to 0} \, \frac{-sin(x)}{cos(x)} \\[5pt] &= \lim_{x\to 0} -tan(x) = 0 \end{align}$$

But from the right, the limit would be infinite.

$$ \begin{align} \lim_{x\to\infty} \frac{e^x + 1}{x^2} \; &\longrightarrow \left( \frac{\infty}{\infty} \right) \\[5pt] \lim_{x\to \infty} \frac{e^x}{2x} &= \lim_{x\to\infty} \, \frac{e^x}{2} \\[5pt] &= \frac{1}{2}\lim_{x\to \infty} \, e^x \; \rightarrow \; = \infty \end{align}$$

$$\lim_{x\to\ 0^-} \, x \, e^{\frac{1}{x}}$$

We have to be careful with this function. Take a look at its graph:

Notice that the limits at x = 0 are different from the left and the right. We can work it out using L'Hopital's rule. First manipulate it into a ∞/∞ limit like this:

$$\lim_{x\to 0^-} \, \frac{e^{1/x}}{\frac{1}{x}}$$

$$\lim_{x\to 0^-} \, \frac{e^{1/x}}{\frac{1}{x}} \; \longrightarrow \left( \frac{\infty}{\infty} \right)$$

Now when x is positive, we have:

$$ \require{cancel} \lim_{x\to 0^-} \frac{e^{1/x} \cancel{\left( \frac{-1}{x^2} \right)}}{\cancel{\frac{-1}{x^2}}} = \lim_{x\to 0^-} \, e^{1/x} = e^{\infty} \; \rightarrow \; \infty$$

But when x < 0, that is, when we approach zero from the left, we have:

$$\lim_{x\to 0} \frac{1}{e^{\frac{1}{x}}} = 0$$

$$ \begin{align} \lim_{x\to 0} \, \frac{1 - cos(x)}{x^2} \; &\longrightarrow \left( \frac{0}{0} \right) \\[5pt] \lim_{x\to 0} \frac{sin(x)}{2x} &\longrightarrow \left( \frac{0}{0} \right) \\[5pt] \lim_{x\to 0} \frac{cos(x)}{2} &= \bf{\frac{1}{2}} \end{align}$$

$$ \begin{align} \lim_{x\to 0} \, x \, csc(x) = \lim_{x\to 0} \, \frac{x}{sin(x)} \; &\longrightarrow \left( \frac{0}{0} \right) \\[5pt] \lim_{x\to 0} \, \frac{1}{cos(x)} &= 1 \end{align}$$

$$ \begin{align} \lim_{x\to 1} \, \frac{x^3 - x^2}{x^3 - 1} \; &\longrightarrow \; \left( \frac{0}{0} \right) \\[5pt] &= \lim_{x\to 1} \, \frac{3x^2 - 2x}{3x^2} \\[5pt] &= \frac{3 - 2}{3} = \frac{1}{3} \end{align}$$

$$ \begin{align} \lim_{x\to 0} \, \frac{tan(x)}{tan(3x)} \; &\longrightarrow \; \left( \frac{0}{0} \right) \\[5pt] &= \lim_{x\to 0} \, \frac{sec^2(x)}{3 sec^2(3x)} = \lim_{x\to 0} \, \frac{\frac{1}{cos^2(x)}}{\frac{3}{cos^2(3x)}} \\[5pt] &= \lim_{x\to 0} \, \frac{cos^2(3x)}{3 cos^2(x)} = \frac{cos^2(0)}{3 cos^2(0)} = \frac{1}{3} \end{align}$$

$$ \begin{align} \lim_{x\to\infty} \, x^3 e^{-x} &= \lim_{x\to\infty} \frac{x^3}{e^x} \; \longrightarrow \; \left( \frac{\infty}{\infty} \right) \\[5pt] &= \lim_{x\to\infty} \frac{3x^2}{e^x} = \lim_{x\to\infty} \frac{6x}{e^x} \\[5pt] &= \lim_{x\to\infty} \frac{6}{e^x} = 0 \end{align}$$

$$\lim_{x\to 0} \, x^2 \, ln(x)$$

We can manipulate this limit into an $\frac{\infty}{\infty}$ limit like this:

$$ \begin{align} \lim_{x\to 0} \, \frac{ln(x)}{\frac{1}{x^2}} \; &\longrightarrow \; \left( \frac{\infty}{\infty} \right) \\[5pt] &= \lim_{x\to 0} \frac{\frac{1}{x}}{\frac{-2}{x^3}} = \lim_{x\to 0} \frac{-x^3}{2x} \\[5pt] &= \lim_{x\to 0} \frac{-x^2}{2} = 0 \end{align}$$

$$\lim_{x\to 0^+} \, x^{sin(x)}$$

This one is do-able, but we have to use some trickery. Instead, find the limit of:

$$\lim_{x\to 0^+} \, ln\left( x^{sin(x)} \right) = \lim_{x\to 0^+} \, sin(x) ln(x)$$

Using the properties of limits, we'll be able to take the result of this as an exponent of e to get the final answer. Now re-express as:

$$ \begin{align} \lim_{x\to 0^+} \, \frac{ln(x)}{\frac{1}{sin(x)}} \; &\longrightarrow \; \left( \frac{\infty}{\infty} \right) \\[5pt] &= \lim_{x\to 0^+} \frac{\frac{1}{x}}{\frac{cos(x)}{sin^2(x)}} \\[5pt] &= \lim_{x\to 0^+} \frac{sin^2(x)}{x cos(x)} \; \longrightarrow \; \left( \frac{0}{0} \right) \\[5pt] &= \lim_{x\to 0^+} \frac{2 sin(x) cos(x)}{cos(x) - x sin(x)} = \frac{0}{1} = 0 \end{align}$$

Now we actually found the log (ln) of the limit, so we have to undo that:

$$\lim_{x\to 0^+} \, x^{sin(x)} = e^0 = 1$$

This is a fascinating limit. A first look, particularly in the context of a L'Hopital's rule section, is that it's an ∞/∞ limit, so we can apply L'Hopital's rule like this:

$$ \begin{align} \lim_{x\to\infty} \, \frac{\sqrt{1 + x^2}}{x} \; &\longrightarrow \; \left( \frac{\infty}{\infty} \right) \\[5pt] &= \lim_{x\to\infty} \frac{\frac{1}{\cancel{2}}(1 + x^2)^{-1/2}\cancel{2}x}{1} \\[5pt] &= \lim_{x\to\infty} \frac{x}{\sqrt{1 + x^2}} \end{align}$$

Now this is just the reciprocal of our original limit, and a little thought will convince you that if we apply L'Hopital's rule again, we'll just get back to the original limit; it's a never-ending loop. So how to solve it? Using the properties of limits, try squaring the whole limit:

$$\left( \lim_{x\to \infty} \frac{\sqrt{1 + x^2}}{x} \right)^2 = \lim_{x\to \infty} \frac{1 + x^2}{x^2} = 1.$$

Now if we just take the square root of that result, the solution is 1.

$$\lim_{x\to\infty} \, (e^{-x})^{\frac{1}{x}} = \lim_{x\to\infty} \left( \frac{1}{e^x} \right)^{\frac{1}{x}}$$

This is a $0^0$ limit. We can solve it by taking the limit of the natural log of this function, remembering that we'll have to raise our result as a power of e to finish:

$$ \begin{align} &\lim_{x\to\infty} \, ln\left[ (e^{-x})^{\frac{1}{x}} \right] \\[5pt] &= \lim_{x\to\infty} \, \frac{1}{x} \, ln(e^{-x}) = \lim_{x\to\infty}\, \frac{-x}{x} = -1 \end{align}$$

Now raise that result as a power of e:

$$e^{-1} = \frac{1}{e}$$

$$ \begin{align} \lim_{x\to\infty} \, \frac{e^x}{x} \; &\longrightarrow \; \left( \frac{\infty}{\infty} \right) \\[5pt] &= \lim_{x\to\infty} \frac{e^x}{1} \; \longrightarrow \; \infty \end{align}$$

This makes sense. The exponential function in the numerator grows faster than the polynomial function in the denominator.