Q.Evaluate the following integrals:

At $x=2$: numerator $=8-4-2-2=0$ and denominator $=16-12-6+2=0$, so it is $\tfrac00$. Factor out $(x-2)$:

Hence …

The key insight here is that the expression is a sum of two separate pieces, each with its own behaviour as $x \to 2$. You cannot just plug $x=2$ into the whole thing — the cosine term oscillates wildly, and the rational part has a removable singularity. The trick is to handle each term independently, using the squeeze theorem for the first and algebraic simplification for the second.

1. First term: $(x^2 - 4x + 4) \cos\left(\frac{2}{x-2}\right)$ Notice $x^2 - 4x + 4 = (x-2)^2$. So the first term is $(x-2)^2 \cos\left(\frac{2}{x-2}\right)$. As $x \to 2$, $(x-2)^2 \to 0$. The cosine factor, no matter how wildly it oscillates, is always between $-1$ and $1$ — it is bounded. By the squeeze theorem:

Both the left and right bounds go to $0$ as $x \to 2$, so the first term tends to $0$.

2. Second term: $\frac{x^2 - 4}{x^3 - 2x - 4}$ This is a rational expression. First, factor the numerator: $x^2 - 4 = (x-2)(x+2)$. The denominator is a cubic. Check if $x=2$ is a root: $2^3 - 2(2) - 4 = 8 - 4 - 4 = 0$, so $x-2$ is a factor. Perform polynomial division or factor by grouping: $x^3 - 2x - 4 = (x-2)(x^2 + 2x + 2)$ …

When $\theta$ approaches $\frac{\pi}{2}$ from the left, $\tan\theta$ grows without bound — it shoots off to $+\infty$. So instead of worrying about the angle directly, we can let $x = \tan\theta$ and consider $x \to \infty$. This turns a trigonometric limit into a rational function limit, which is much easier to handle.

The expression becomes:

Now, both numerator and denominator are polynomials. As $x \to \infty$, the highest-degree terms dominate. So we compare the leading terms.

1. Expand the denominator

   $(3 - 2x)^4 = ( -2x + 3)^4$. The leading term comes from $(-2x)^4 = 16x^4$. The lower-degree terms (like $(-2x)^3 \cdot 3$, etc.) are negligible for large $x$.

2. Compare leading terms

   Numerator leading term: $8x^4$.

   Denominator leading term: $16x^4$.

   So the limit is:

3. Check for sign issues …

Concept & Intuition

When you plug $x=3$ directly into the fraction, both numerator and denominator become zero. That’s a classic $\frac{0}{0}$ indeterminate form — it doesn’t tell us the limit yet. The trick is that $x=3$ makes both polynomials vanish, so $(x-3)$ must be a factor of each. By factoring and cancelling that common factor, we remove the “hole” and get a simpler expression whose value at $x=3$ is the limit.

Step-by-step solution

1. Check direct substitution

   Let $N(x)=x^3-3x^2-4x+12$ and $D(x)=2x^3-7x^2+2x+3$.

   At $x=3$:

   $N(3)=27-27-12+12=0$

   $D(3)=54-63+6+3=0$

   So we have $\frac{0}{0}$ — proceed to factoring.

2. Factor the numerator

   Since $x=3$ is a root, $(x-3)$ divides $N(x)$. Use synthetic division (or polynomial long division) with root $3$:

   Coefficients: $1,\,-3,\,-4,\,12$

   Bring down $1$; multiply by $3$ → $3$, add to $-3$ → $0$; multiply $0$ by $3$ → $0$, add to $-4$ → $-4$; multiply $-4$ by $3$ → $-12$, add to $12$ → $0$.

   The quotient is $x^2+0x-4 = x^2-4$.

   So $N(x)=(x-3)(x^2-4) = (x-3)(x-2)(x+2)$.

3. Factor the denominator

   Check $D(3)=0$, so $(x-3)$ is also a factor. Synthetic division with $3$ on coefficients $2,\,-7,\,2,\,3$:

   Bring down $2$; $2\cdot3=6$, add to $-7$ → $-1$; $-1\cdot3=-3$, add to $2$ → $-1$; $-1\cdot3=-3$, add to $3$ → $0$.

   The quotient is $2x^2 - x - 1$.

   Factor $2x^2 - x - 1 = (2x+1)(x-1)$.

   So $D(x)=(x-3)(2x+1)(x-1)$.

4. Cancel the common factor

   For $x \neq 3$,

   $$$$

Direct substitution of $x=1$ gives $\dfrac00$, so a common factor must cancel.

1. Factor the denominator.

2. Handle the numerator root. For the option list to be finite the vanishing factor must be first order in $(x-1)$, i.e. the numerator carries $\sqrt x-1$, and

3. Cancel and substitute. …

We want

As $x \to 1$, $1-x \to 0$ and $\tan(\pi x/2) \to \tan(\pi/2)$, which blows up to infinity. So we have a $0 \cdot \infty$ form — not directly evaluable. The trick is to rewrite the tangent in terms of a cotangent, because near $\pi/2$ the tangent behaves like the reciprocal of a small quantity.

Why this works:

Recall $\tan\theta = \cot(\pi/2 - \theta)$. When $\theta$ is near $\pi/2$, the argument $\pi/2 - \theta$ is small, and $\cot$ of a small angle behaves like $1/(\text{small angle})$. This turns the product into a ratio, which is a standard limit.

1. Rewrite the tangent Let $\theta = \frac{\pi}{2} x$. Then

So the limit becomes

2. Introduce a new variable Set $t = 1-x$. Then as $x \to 1$, $t \to 0^+$. The limit is

3. Express cotangent as cosine over sine

So the product becomes

4. Use the small-angle limit We know $\displaystyle \lim_{u \to 0} \frac{\sin u}{u} = 1$. Here $u = \frac{\pi}{2} t$, so $\frac{t}{\sin(\pi t/2)} = \frac{t}{\frac{\pi}{2} t \cdot \frac{\sin(\pi t/2)}{\pi t/2}} = \frac{1}{\frac{\pi}{2} \cdot \frac{\sin(\pi t/2)}{\pi t/2}}.$ …

We want

The numerator is a difference of exponentials with the same base. The key idea: factor out $2^{\sin x}$ and then use the series expansion for $e^u$ when $u$ is small. This turns a tricky limit into a simple algebraic comparison of $\tan x$ and $\sin x$ near zero.

1. Factor and rewrite

As $x \to 0$, $\sin x \to 0$, so $2^{\sin x} \to 1$. Also, $\tan x - \sin x \to 0$, so we can use $2^u - 1 = e^{u\log 2} - 1 \sim (u \log 2)$ for small $u$.

2. Approximate the difference For small $u$, $e^{u\log 2} - 1 \sim u \log 2$. Hence

Since $2^{\sin x} \to 1$, the leading behavior is

3. Expand $\tan x - \sin x$ Recall:

Subtract:

4. Plug into the limit …

The core idea here is that as $x$ approaches $1$, both the numerator $\log x$ and the denominator $1-x$ approach $0$. This gives the $\frac{0}{0}$ indeterminate form, which means we cannot just substitute $x=1$ — we need to rewrite the expression into a form whose limit is known.

The natural logarithm has a beautiful property near $1$: for small $t$, $\log(1+t) \approx t$, and more precisely, $\lim_{t \to 0} \frac{\log(1+t)}{t} = 1$. This is a standard result derived from the definition of the derivative of $\log x$ at $x=1$, and it is the key to unlocking this problem.

1. Substitute to create a small variable. Let $t = x - 1$. Then as $x \to 1$, we have $t \to 0$. Also, $x = 1 + t$, so $\log x = \log(1+t)$ and $1-x = 1 - (1+t) = -t$. The limit becomes:

2. Factor out the constant. The denominator $-t$ can be written as $(-1) \cdot t$, so:

3. Apply the standard limit. We know that $\lim_{t \to 0} \frac{\log(1+t)}{t} = 1$. This is a fundamental limit — you can prove it using the series expansion $\log(1+t) = t - \frac{t^2}{2} + \cdots$ or by recognizing it as the derivative of $\log x$ at $x=1$: $$$$