Some important other limits

Result 9.2. $\displaystyle \lim_{x\to0}\frac{e^x-1}{x}=1.$

This result is used here as a known building block (a full elementary justification needs either the series definition of $e^x$ or tools developed in a later chapter, so, in keeping with the book, it is quoted rather than re-derived from first principles).

Result 9.3. $\displaystyle \lim_{x\to0}\frac{a^x-1}{x}=\log_e a,\quad a>0.$

Proof. Since $a^x$ and $\log_a x$ are inverse functions of each other, $a^x$ can always be rewritten in terms of the natural exponential as $a^x=e^{x\log_e a}$. Hence

Substitute $y=x\log_e a$; as $x\to0$, $y\to0$ as well, and the second factor becomes exactly $\dfrac{e^y-1}{y}$, whose limit as $y\to0$ is $1$ by Result 9.2. So the whole expression tends to $\log_e a\cdot1=\log_e a$. $\blacksquare$

Result 9.4. $\displaystyle \lim_{x\to0}\frac{\log(1+x)}{x}=1.$

Proof. Set $y=\log(1+x)$, so $y\to0$ as $x\to0$, and $1+x=e^y$, i.e. $x=e^y-1$. Substituting,

and as $y\to0$ the denominator tends to $1$ by Result 9.2, so the whole fraction tends to $\dfrac11=1$. $\blacksquare$

Results 9.5–9.9 (stated without proof, used freely from here on):

The number $e$ that keeps appearing here is a transcendental number — it never satisfies any polynomial equation with rational coefficients, $a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0=0$ — which is part of why it cannot be pinned down by ordinary algebra and is instead defined through a limit such as Result 9.7.

Worked techniques (Examples 9.32–9.36).

• Reciprocal substitution into the $e$-form. For a limit such as $\lim_{x\to0}(1+\sin x)^{2\csc x}$, substitute $y=1/\sin x$ (so $y\to\infty$ as $x\to0$) to rewrite the whole expression as $\left(1+\dfrac1y\right)^{2y}$ — exactly the Result 9.7/9.9 pattern — which evaluates to $e^2$.
• Splitting a $(1+\cdot)^{\text{power}}$ limit at infinity. For $\lim_{x\to\infty}\left(\dfrac{x+2}{x-2}\right)^{x}$, first rewrite the base as $1+\dfrac{4}{x-2}$, substitute $y=\dfrac{x-2}{4}$ (so $y\to\infty$ as $x\to\infty$, and $x=4y+2$), and split the resulting exponent into a "$4y$" part — which reduces to the standard $e^4$ form — and a leftover finite-exponent part that itself tends to $1$; the two combine to give $e^4$.
• Reducing a trigonometric-power limit to the Result 9.7 pattern. For $\lim_{x\to\pi/4}\dfrac{(\cos x-\sin x)^5}{(1-\sin2x)^{5/2}}$, use the identity $(\cos x-\sin x)^2=1-\sin2x$ to rewrite both numerator and denominator purely in terms of $(1-\sin2x)$, collapsing the whole quotient to a single power of that quantity; substituting $y=1+\sin2x$ (so $y\to2$ as $x\to\pi/4$) then finishes the limit by direct substitution. …