Q.If $f(x) = x^2$, find $\dfrac{f(1.1) - f(1)}{(1.1 - 1)}$.

Concept and Intuition

Whenever an expression looks like $f(a+\delta(h))-f(a)$ with $\delta(h)\to0$ as $h\to0$, we can write it as $\dfrac{f(a+\delta)-f(a)}{\delta}\cdot\delta \to f'(a)\cdot\delta(h)$ for small $h$. Applying this to both numerator and denominator turns the whole limit into a ratio of derivatives times a ratio of the two $\delta$'s.

Step-by-Step Solution

1. Numerator argument: $2h+2+h^2\to 2$ as $h\to0$, with $\delta_1=2h+h^2$.

2. Denominator argument: $h-h^2+1\to1$ as $h\to0$, with $\delta_2=h-h^2$.

3. Ratio: $\dfrac{6(2h+h^2)}{4(h-h^2)}=\dfrac{6h(2+h)}{4h(1-h)}=\dfrac{6(2+h)}{4(1-h)}$.
4. Take $h\to0$: $\dfrac{6(2)}{4(1)}=\dfrac{12}{4}=3$.

Common Mistakes

• Using $f'(1)$ for the numerator or $f'(2)$ for the denominator (swapping them).
• Forgetting to factor out the common $h$ before taking the limit, leading to a $0/0$ dead end.

ANSWER: D

Concept and Intuition

When a piecewise function's formula involves $e^{1/x}$-type terms, the behaviour as $x\to0^+$ and $x\to0^-$ can be drastically different because $e^{1/x}\to\infty$ from one side and $\to0$ from the other. Checking differentiability then means computing $\lim\frac{f(x)-f(0)}{x}$ separately from each side.

Step-by-Step Solution

1. $f(0)=0$, so $f'(0^\pm)=\displaystyle\lim_{x\to0^\pm}\frac{f(x)}{x}=\lim_{x\to0^\pm}\frac{2e^{1/(2x)}-3e^{-1/(2x)}}{e^{1/(2x)}+4e^{-1/(2x)}}$.
2. As $x\to0^+$: $1/(2x)\to+\infty$, so $e^{1/(2x)}\to\infty$, $e^{-1/(2x)}\to0$. Divide numerator and denominator by $e^{1/(2x)}$: $\dfrac{2-3e^{-1/x}}{1+4e^{-1/x}}\to\dfrac{2-0}{1+0}=2$. So $f'(0^+)=2$.
3. As $x\to0^-$: $1/(2x)\to-\infty$, so $e^{1/(2x)}\to0$, $e^{-1/(2x)}\to\infty$. Divide by $e^{-1/(2x)}$: $\dfrac{2e^{1/x}-3}{e^{1/x}+4}\to\dfrac{0-3}{0+4}=-\dfrac34$. So $f'(0^-)=-\dfrac34$.
4. Since $f'(0^+)=2\ne-\dfrac34=f'(0^-)$, the derivative at $0$ does not exist — $f$ is not differentiable at $x=0$.

Common Mistakes

• Dividing by the wrong exponential term on each side (e.g. always dividing by $e^{1/(2x)}$ regardless of the sign of $x$), which gives indeterminate $\infty/\infty$ or wrong values.
• Assuming continuity of $f$ implies differentiability.

ANSWER: C

Concept and Intuition

Whenever a limit expression contains $\sin(\text{something}\to\pm\infty)$, the sine keeps oscillating between $-1$ and $1$ indefinitely and never settles on one value, so such a limit fails to exist (unless it's multiplied by something that forces it to zero, e.g. $x\sin(1/x)\to0$, which is not the case here since there's no vanishing factor).

Step-by-Step Solution

1. For $x\neq0$: $f(x)=x\left(1+\dfrac12\sin(\log x^2)\right)$, and $f(0)=0$.
2. $\dfrac{f(x)-f(0)}{x}=\dfrac{f(x)}{x}=1+\dfrac12\sin(\log x^2)$.
3. As $x\to0$, $x^2\to0^+$, so $\log x^2\to-\infty$.
4. $\sin(\log x^2)$ therefore oscillates between $-1$ and $1$ infinitely often as $x\to0$ (it never approaches a single value) — there is no factor forcing this oscillation to damp to $0$.
5. Hence $1+\dfrac12\sin(\log x^2)$ also oscillates (between $1/2$ and $3/2$) and the limit does not exist.

Common Mistakes

• Assuming the bounded factor $\frac12\sin(\cdot)$ automatically vanishes in the limit — that only happens if it's multiplied by something tending to $0$ (like the "$x\sin(1/x)$" trick), which isn't the structure here.
• Trying to assign the limit a specific value like $f(0)$ or $1/2$ without checking that the oscillation genuinely never settles.

ANSWER: B

Concept and Intuition

The trick is to rewrite $xf(m)-mf(x)$ so that a genuine difference quotient $\dfrac{f(x)-f(m)}{x-m}\to f'(m)$ appears — add and subtract $mf(m)$.

Step-by-Step Solution

1. $xf(m)-mf(x) = xf(m)-mf(m)+mf(m)-mf(x) = f(m)(x-m) - m\big(f(x)-f(m)\big)$.
2. So $\dfrac{xf(m)-mf(x)}{x-m} = f(m) - m\cdot\dfrac{f(x)-f(m)}{x-m}$.
3. Taking $x\to m$: the limit is $f(m) - m f'(m)$.
4. The given equation becomes: $\big[f(m)-mf'(m)\big] + f'(m) = f(m)$.
5. Simplify: $-mf'(m)+f'(m)=0 \Rightarrow f'(m)(1-m)=0$.
6. Since $f'(m)\ne0$ is given, we must have $1-m=0$, so $m=1$.

Common Mistakes

• Trying to apply L'Hôpital blindly to $xf(m)-mf(x)$ over $x-m$ without recognizing it's not automatically a $0/0$ form in the usual sense (it needs the algebraic split shown above, since it's not literally L'Hôpital's setup with a fixed function of $x$ alone in both places).
• Sign errors when distributing the minus sign through $-m(f(x)-f(m))$.

ANSWER: C

Concept and Intuition

The bracketed limit has the shape $\displaystyle\lim_{h\to0}\dfrac{g(x)-g(x+h)}{h}$ which is $-g'(x)$ for $g(t)=1/t$ evaluated appropriately — but it's simplest to just combine the fraction directly and cancel the $(y-2)$ factor before taking the limit.

Step-by-Step Solution

1. Let $u=y-2$, so $u\to0$ as $y\to2$.
2. $\dfrac1u\left(\dfrac1x-\dfrac1{x+u}\right) = \dfrac1u\cdot\dfrac{(x+u)-x}{x(x+u)} = \dfrac1u\cdot\dfrac{u}{x(x+u)} = \dfrac1{x(x+u)}$.
3. Taking $u\to0$: $\dfrac1{x\cdot x} = \dfrac1{x^2}$.
4. Now differentiate with respect to $x$: $\dfrac{d}{dx}\left(x^{-2}\right) = -2x^{-3} = -\dfrac2{x^3}$.

Common Mistakes

• Trying to take the limit before combining the fractions, which leads to a $0/0$ indeterminate form handled clumsily.
• Sign error when differentiating $x^{-2}$.

ANSWER: C

Concept and Intuition

Since $f''$ is continuous at $0$, $f$ has a second-order Taylor expansion there: $f(x)\approx f(0)+f'(0)x+\frac{f''(0)}{2}x^2$. Plugging $x,2x,4x$ into this and combining lets the constant and linear terms cancel by design (a hallmark of these limit problems), leaving a pure multiple of $f''(0)$.

Step-by-Step Solution

1. $f(x)\approx f(0)+f'(0)x+\frac{f''(0)}{2}x^2$
2. $f(2x)\approx f(0)+2f'(0)x+2f''(0)x^2$
3. $f(4x)\approx f(0)+4f'(0)x+8f''(0)x^2$
4. $2f(x)-3f(2x)+f(4x)$: constant terms $2f(0)-3f(0)+f(0)=0$; linear terms $[2-6+4]f'(0)x=0$; quadratic terms $\left[2\cdot\frac12-3\cdot2+8\right]f''(0)x^2=3f''(0)x^2$.
5. So the limit $=3f''(0)=3(4)=12$.

Common Mistakes

• Stopping the Taylor expansion at first order and missing the needed $x^2$ term entirely.
• Arithmetic slip in the coefficient $2(\tfrac12)-3(2)+8=1-6+8=3$.

ANSWER: C

Concept and Intuition

A constant derivative means $f$ must be a straight line $f(x)=mx+c$; here $m=14$. Knowing one point, $f(2)=14$, pins down $c$ completely.

Step-by-Step Solution

1. $f'(x)=14$ (constant) $\Rightarrow f(x)=14x+c$.
2. $f(2)=14 \Rightarrow 28+c=14 \Rightarrow c=-14$. So $f(x)=14x-14$.
3. $xf(2)-2f(x) = 14x - 2(14x-14) = 14x-28x+28=-14x+28=-14(x-2)$.
4. $\dfrac{xf(2)-2f(x)}{x-2}=\dfrac{-14(x-2)}{x-2}=-14$ for all $x\ne2$, so the limit as $x\to2$ is simply $-14$.

Common Mistakes

• Trying to apply L'Hôpital's rule unnecessarily instead of noticing the exact algebraic cancellation.
• Sign error while distributing $-2f(x)$.

ANSWER: C

Concept and Intuition

As $x\to a$, both $\sqrt{f(x)}\to\sqrt{f(a)}=a$ and $\sqrt{\phi(x)}\to\sqrt{\phi(a)}=b$, so numerator and denominator both vanish — an indeterminate $0/0$ form suited to L'Hôpital's rule, differentiating each under the square root via the chain rule.

Step-by-Step Solution

1. Numerator derivative: $\dfrac{d}{dx}\sqrt{f(x)} = \dfrac{f'(x)}{2\sqrt{f(x)}}$, evaluated at $x=a$: $\dfrac{f'(a)}{2a}$ (using $\sqrt{f(a)}=a$).
2. Denominator derivative: $\dfrac{d}{dx}\sqrt{\phi(x)}=\dfrac{\phi'(x)}{2\sqrt{\phi(x)}}$, at $x=a$: $\dfrac{\phi'(a)}{2b}$.
3. By L'Hôpital: $\lim_{x\to a}\dfrac{\sqrt{f(x)}-a}{\sqrt{\phi(x)}-b} = \dfrac{f'(a)/(2a)}{\phi'(a)/(2b)} = \dfrac{f'(a)\,b}{a\,\phi'(a)}$.
4. Substitute $f'(a)=3\phi'(a)$: $=\dfrac{3\phi'(a)\cdot b}{a\,\phi'(a)}=\dfrac{3b}{a}$.

Common Mistakes

• Forgetting the factor of 2 from differentiating the square root (chain rule).
• Not substituting the given relation $f'(a)=3\phi'(a)$ to eliminate the derivatives from the final answer.

ANSWER: D