Limit flash cards

1. 1.State the standard limit $\lim_{x\to 0}\dfrac{\sin x}{x}$ and $\lim_{x\to 0}\dfrac{\tan x}{x}$.

   Both equal $1$: $\lim_{x\to 0}\dfrac{\sin x}{x}=1$ and $\lim_{x\to 0}\dfrac{\tan x}{x}=1$ (with $x$ in radians).

   Hint: Squeeze theorem on the unit circle.

2. 2.What is the value of $\lim_{x\to 0}\dfrac{e^x-1}{x}$ and $\lim_{x\to 0}\dfrac{\ln(1+x)}{x}$?

   Both limits equal $1$; these follow from the Taylor expansions $e^x\approx 1+x$ and $\ln(1+x)\approx x$ for small $x$.

   Hint: Compare with first-order series expansion.

3. 3.Give the definition of $f(x)$ being continuous at $x=a$.

   $f$ is continuous at $a$ if $\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)$, i.e. the left-hand limit, right-hand limit and function value all exist and are equal.

   Hint: Three-part condition: LHL, RHL, value.

4. 4.State the relation between differentiability and continuity of a function at a point.

   If $f$ is differentiable at $x=a$ it must be continuous there, but continuity does not imply differentiability (e.g. $f(x)=|x|$ is continuous but not differentiable at $x=0$).

   Hint: One-directional implication only.

5. 5.What does L'Hôpital's Rule state for evaluating $\lim_{x\to a}\dfrac{f(x)}{g(x)}$ when it gives $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$?

   If $f,g$ are differentiable near $a$ (except possibly at $a$) with $g'(x)\neq 0$, then $\lim_{x\to a}\dfrac{f(x)}{g(x)}=\lim_{x\to a}\dfrac{f'(x)}{g'(x)}$, provided the latter limit exists (finite or infinite).

   Hint: Differentiate numerator and denominator separately, not as a quotient.

6. 6.Evaluate $\lim_{x\to 0}\dfrac{x-\sin x}{x^3}$.

   This is $\frac{0}{0}$; apply L'Hopital's rule three times: $\dfrac{1-\cos x}{3x^2}\to\dfrac{\sin x}{6x}\to\dfrac{\cos x}{6}$. As $x\to0$, $\cos x\to1$, so the limit is $\dfrac{1}{6}$.

   Hint: Differentiate repeatedly; it stays $0/0$ each time.

7. 7.Evaluate $\lim_{x\to 0}\dfrac{\tan x-x}{x^3}$.

   By L'Hopital, $\dfrac{\sec^2x-1}{3x^2}=\dfrac{\tan^2x}{3x^2}=\dfrac13\left(\dfrac{\tan x}{x}\right)^2$. Since $\dfrac{\tan x}{x}\to1$, the limit equals $\dfrac13$.

   Hint: Rewrite $\sec^2x-1$ as $\tan^2x$ before taking the next limit.

8. 8.Evaluate $\lim_{x\to \pi/2}(\sec x-\tan x)$.

   Write $\sec x-\tan x=\dfrac{1-\sin x}{\cos x}$, which is $\frac00$ at $x=\pi/2$. L'Hopital gives $\dfrac{-\cos x}{-\sin x}=\dfrac{\cos x}{\sin x}\to\dfrac{0}{1}$. So the limit is $0$.

   Hint: Combine into a single fraction first; then differentiate.

9. 9.Evaluate $\lim_{x\to 0}\dfrac{xe^x-\ln(1+x)}{x^2}$.

   This is $\frac00$. Differentiating once gives $\dfrac{e^x+xe^x-\frac1{1+x}}{2x}$, still $\frac00$ at $x=0$ (value $0$). Differentiating again gives $\dfrac{2e^x+xe^x+\frac1{(1+x)^2}}{2}\to\dfrac{2+0+1}{2}=\dfrac32$ as $x\to0$. So the limit is $\dfrac32$.

   Hint: You will need L'Hopital's rule twice, not once.

10. 10.Evaluate $\lim_{x\to 0}\dfrac{\sin x-x\cos x}{x^3}$.

    This is $\frac00$; L'Hopital gives $\dfrac{\cos x-\cos x+x\sin x}{3x^2}=\dfrac{x\sin x}{3x^2}=\dfrac13\cdot\dfrac{\sin x}{x}\to\dfrac13\cdot1$. So the limit is $\dfrac13$.

    Hint: Differentiate the numerator carefully using the product rule on $x\cos x$.

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