Special limits summary
Special trig limits
There are two commonly used special Calculus limits involving trig functions. They are:
$$\lim_{x\to 0} \frac{\sin x}{x} =1$$
and
$$\lim_{x\to 0} \frac{\cos x – 1}{x} = 0.$$
These limit are necessary for determining the derivatives of $\sin x$ and $\cos x$ and can both be proved geometrically. You can see the details of the proofs further down if you’re interested. For now, let’s look at how to solve limit problems using these limits.
Example
What is $\displaystyle \lim_{x\to 0} \frac{\cos(3x)-1}{x}?$
Solution
We can start by recognizing that this looks related to the limit
$$\lim_{x\to 0} \frac{\cos x -1 }{x}.$$
This difference is that there is a $3x$ inside the cosine function. The easiest way to deal with this is to use a change of variable. If we take $t = 3x,$ then $t\to 0$ as $x\to 0$ and $x = \frac{s}{3}$, so the limit can be rewritten as follows:
$$
\begin{aligned}
\lim_{x\to 0} \frac{\cos(3x)-1}{x} &= \lim_{t\to 0}3 \frac{\cos(t)-1}{ t} \\
& = 3 \lim_{t\to 0} \frac{\cos t}{t}\\
& = 0.
\end{aligned}
$$
Example
What is $\displaystyle \lim_{x\to 0} \frac{x}{\sin(2x)}?$
Solution
We can start by recognizing that this looks related to the limit
$$\lim_{x\to 0} \frac{\sin x }{x}.$$
In fact using limit rules we can say that
$$
\lim_{x\to 0} \frac{x}{\sin x} = \frac{1}{\lim_{x\to 0} \frac{\sin x}{x}} = \frac{1}{1} = 1.
$$
The problem now is that we have a $2x$ inside the sine function. The easiest way to deal with this is to use a change of variable. If we take $s = 2x,$ then $s\to 0$ as $x\to 0$ and $x = \frac{s}{2}$, so the limit can be rewritten as follows:
$$
\begin{aligned}
\lim_{x\to 0} \frac{x}{\sin(2x)} &= \lim_{s\to 0} \frac{s}{2 \sin (s)} \\
& = \frac{1}{2} \lim_{s\to 0} \frac{s}{\sin (s)}\\
& = \frac{1}{2}.
\end{aligned}
$$
Now you try it!
Practice Problems for special limits in Calculus
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Special limits with exponentials
The special limit involving exponential functions that we see in Calculus is
$$
\lim_{x\to 0} \frac{e^x-1}{x} = 1. \tag{1}
$$
This limit is interesting because it can actually be used as the definition of $e^x$. By this I mean we can define $e^x$ the the unique exponential function with positive base that satisfies this limit. We can also define $e^x$ in quite a few other ways, including with another limit:
$$
e^x = \lim_{n\to \infty} \left(1+ \frac{x}{n}\right)^n.
$$
We’ll discuss a few other definitions and how they connect to this first limit a bit later in the page, but for now let’s practice evaluating limits that are related to (1).
Example
What is $\displaystyle \lim_{x\to 0} \frac{2^x -1}{x}?$
Solution
We can see right away that this limit seems related to the limit
$$
\lim_{x\to 0} \frac{e^x-1}{x} = 1.
$$
To evaluate it we need to try and rewrite $2^x$ in terms of $e^x$. To do this we can use the trick that $\ln x$ and $e^x$ are inverses, so
$$
2^x = e^{\ln (2^x)}.
$$
Then using log rules this can be rewritten as
$$
2^x = e^{x \ln 2}.
$$
This means that similarly to the trig limits we looked at before we have almost our original limit, but with $cx$ instead of $x$. To identify the limit we can do the substitution $t = x \ln 2$ and $x = \frac{t}{\ln 2}$. Therefore
$$
\begin{aligned}
\lim_{x\to 0} \frac{2^x -1}{x} & = \lim_{t\to 0} \frac{e^t-1}{\frac{t}{\ln 2}}\\
& = \ln 2 \cdot \lim_{t\to 0} \frac{e^t -1}{t} = \ln 2.
\end{aligned}
$$
Special limits with logarithms
The special limit involving exponential functions that we see in Calculus is
$$
\lim_{x\to 0} \frac{\ln(1+x)}{x} = 1. \tag{1}
$$
This limit can be derived a variety of ways. The “correct” way depends on what you’re using for your definition of $\ln x$. Here let’s take a look at how you can see from the limit definition of $e$:
$$
e = \lim_{n\to \infty} \left( 1+ \frac{1}{n}\right)^n.
$$
We can start by rewriting our limit using log rules, and then pulling the limit through the logarithm using the continuity of the logarithm function:
$$
\begin{aligned}
\lim_{x\to 0} \frac{\ln(1+x)}{x} & = \lim_{x\to 0} \ln \left( (1+x)^{1/x}\right)\\
& = \ln \left( \lim_{x\to 0} \big( 1+x\big)^{1/x}\right).
\end{aligned}
$$
Then to finish it off we just take $s=1/x$, in which case $s\to \pm \infty$ as $x\to 0$. We can check that in both cases we have
$$\lim_{s\to \pm \infty} \left(1 + \frac{1}{s}\right)^s = e
$$
and $\ln(e)=1$!
Example
What is $\displaystyle \lim_{x\to 0} \frac{e^{2x} -1}{\ln(1+x)}?$
Solution
The key first step here is to recognize that both bottom and top function would be easier to work with if they were “paired” with $x$. We can do this by multiplying top and bottom by $x$:
$$
\begin{aligned}
\lim_{x\to 0} \frac{e^{2x}-1}{\ln(1+x)} &= \lim_{x\to 0} \frac{e^{2x}-1}{x}\cdot \frac{x}{\ln(1+x)}
\end{aligned}
$$
Splitting these up and working with them separately we can see that
$$
\begin{aligned}
\lim_{x\to 0} \frac{e^{2x} -1}{x} & = \lim_{t\to 0} \frac{e^t-1}{\frac{t}{2}}\\
& = 2 \cdot \lim_{t\to 0} \frac{e^t -1}{t} = 2.
\end{aligned}
$$
where $t = 2x$.
And we can use limit rules on the final one to get:
$$
\begin{aligned}
\lim_{x\to 0} \frac{x}{\ln(1+x)} & = \frac{1}{\lim_{x\to 0} \frac{\ln(1+x)}{x}} \\& = \frac{1}{1} = 1.
\end{aligned}
$$
Together this gives us that the original limit is $2$.
Proof that $\lim_{x\to 0} \frac{\sin x}{x} = 1$
To prove that $\lim_{x\to 0} \frac{\sin x}{x} =1$ we are going to show that
$$
1\ge \frac{\sin x}{x} \ge \cos x.
$$
and then use the squeeze theorem to conclude that the limit of $\frac{\sin x}{x}$ is $1$ as $x\to 0$. Let’s take a look at the squeeze theorem real quick to see why having the inequality above would be enough.
The Squeeze Theorem
Let $f(x) \le g(x) \le h(x)$ around the point $a$. If
$$\lim_{x\to a} f(x)= L = \lim_{x\to a} h(x),$$
Then $\lim_{x\to a} g(x)$ exists and is equal to $L$.
Therefore if we have that
$$\cos x \le \frac{\sin x}{x} \le 1$$
and know
$$
\lim_{x\to 0} \cos x = 1 = \lim_{x\to 0} 1.
$$
then $\lim_{x\to 0} \frac{\sin x}{x} = 1$. So let’s prove that inequality!
We’ll start by checking out some triangles and circle sections and finding their area 🙂
In the first picture we have a triangle with height $\sin \theta$ and base $1$ (assuming this is the unit circle). This means that the area of the triangle is $\frac{1}{2} \sin \theta$.
Now let’s look at the next picture. This is a segment of the circle. The full circle has area $\pi$ and this is $\frac{\theta}{2\pi}$ of the circle (since there are $2\pi$ radians in the full circle), so the area of this wedge is $\pi \frac{\theta}{2\pi} = \frac{\theta}{2}$.
Finally for the last picture we have a triangle $OAC$ (where $O$ is the origin) which is similar to $OBD$. This means that
$$
\frac{1}{\cos \theta} = \frac{|OC|}{|OD|} = \frac{|AC|}{|BD|} = \frac{|AC|}{\sin \theta}.
$$
Rearranging this we get that the height of the triangle $|AC| = \frac{\sin \theta}{\cos \theta} = \tan \theta$.
This means that for $0\lt x\lt \frac{\pi}{2}$
$$
\sin x \le x \le \tan x.
$$
On the same interval we can use algebra to say that
$$
\frac{1}{\sin x} \ge \frac{1}{x} \ge \frac{\cos x}{\sin x}.
$$
Multiplying through by $\sin x$ (which is positive on this interval) we get
$$
1\ge \frac{\sin x}{x} \ge \cos x.
$$
Woot! Woot! Now we just need to extend this to a neighborhood of $0$. But we can do this immediately because all of these functions are even, meaning that plugging in $-x$ gives us the same result. Therefore by the squeeze theorem we have that
$$
\lim_{x\to 0} \frac{\sin x}{x} = 1.
$$
Why is $\lim_{x\to 0} \frac{e^x-1}{x} = 1$?
Why might this limit be one that we care about. Really it seems pretty random, but let’s take a quick adventure into the land of derivatives. Another possible definition of $e^x$ is as the unique function that satisfies that $f'(x) = f(x)$ and $f(0)=1$.
Let’s go ahead an assume that the function $f(x)$ that satisfies this is an exponential function. Proving that this is the case is a bit beyond our scope right now. But if we assume it is exponential, then $f(x) = a^x$ and using the limit definition of the derivative we get:
$$
\begin{aligned}
f'(x) &= \lim_{h\to 0} \frac{a^{x+h}-a^x}{h}\\
& = a^x \ \lim_{h\to 0} \frac{a^h-1}{h}
\end{aligned}
$$
This then returns $a^x$ exactly when $\lim_{h\to 0} \frac{a^h-1}{h} =1$, and we call the $a$ for which this holds $e$!