Let g(x) = (x-1)ⁿ / log cosᵐ (x-1); 0

Q: Wait, why is the LHD of $|x-1|$ at $x=1$ equal to $-1$?

Great question! Remember that for $x < 1$, the expression inside the absolute value, $(x-1)$, is negative. Therefore, $|x-1| = -(x-1) = 1-x$. When you differentiate $1-x$ with respect to $x$, you get $-1$. That is the slope of the left branch of the V-shape.

Q: Why can't we just use L'Hopital's rule immediately?

You *could* use L'Hopital's rule, but it would involve differentiating $\log \cos^m h$ repeatedly, which gets messy with chain rules. Using standard limits is the elegant, 'JEE-approved' way to handle indeterminate forms without getting lost in algebraic weeds.

Q: How do we know $n=2$ just by looking at the power of $h$?

Think about the behavior of $h^k$ as $h \to 0$. If $k > 0$, the limit is $0$. If $k < 0$, the limit is $\infty$. To get a non-zero constant, the power $k$ must be exactly $0$. Here, $k = n-2$, so $n-2=0$ is the only path to a finite, non-zero result.

Question

Let $g(x) = \frac{(x-1)^n}{\log \cos^m (x-1)}; 0 < x < 2, m$ and $n$ are integers, $m eq 0, n > 0$, and let $p$ be the left hand derivative of $|x-1|$ at $x=1$. If $\lim_{x o 1^+} g(x) = p$, then Options: (A) $n=1, m=1$ (B) $n=1, m=-1$ (C) $n=2, m=2$ (D) $n>2, m=n$

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of the V-Shape

Imagine you are standing on a coordinate plane, looking at the graph of $f(x) = |x-1|$. It is a beautiful, sharp V-shape that kisses the x-axis exactly at $x=1$.

That sharp corner is a warning sign in calculus; it tells us the function is non-differentiable there. However, we are looking for the left-hand derivative, $p$.

Since we are approaching $1$ from the left, we are strictly in the region where $x < 1$. In this territory, the absolute value function behaves like $1-x$. The derivative of $1-x$ is simply $-1$.

So, our value for $p$ is locked in: $p = -1$. This is our target.

### The Limit Setup: Shifting the Perspective

Now, we face the main challenge:
$$\lim_{x 	o 1^+} \frac{(x-1)^n}{\log \cos^m (x-1)} = -1$$

Dealing with $x$ approaching $1$ is like trying to solve a puzzle while the pieces are moving. Let us freeze the frame by substituting $x = 1+h$.

As $x$ approaches $1$ from the right, $h$ approaches $0$ from the positive side. Our limit transforms into:
$$\lim_{h 	o 0^+} \frac{h^n}{\log \cos^m h} = -1$$

### The Logarithmic and Trigonometric Dance

We have a power $m$ inside the logarithm. Using the power rule $\log(a^b) = b \log a$, we pull the $m$ out:
$$\lim_{h 	o 0^+} \frac{h^n}{m \log \cos h} = -1$$

Now, look at the denominator. We need to evaluate $\log \cos h$ as $h 	o 0$. We know the standard limit $\lim_{t 	o 0} \frac{\log(1+t)}{t} = 1$.

Let us rewrite $\cos h$ as $1 + (\cos h - 1)$. By multiplying and dividing by $(\cos h - 1)$, we force the expression into the standard form. As $h 	o 0$, the log term effectively becomes $(\cos h - 1)$.

Our expression simplifies to:
$$\lim_{h 	o 0^+} \frac{h^n}{m(\cos h - 1)} = -1$$

### The Final Power Balance

We know that for small $h$, $1 - \cos h \approx \frac{h^2}{2}$. Therefore, $\cos h - 1 \approx -\frac{h^2}{2}$.

Substituting this into our limit, we get:
$$\lim_{h 	o 0^+} \frac{h^n}{m(-\frac{h^2}{2})} = -1$$

Rearranging this, we find:
$$\lim_{h 	o 0^+} \frac{-2 \cdot h^{n-2}}{m} = -1$$

For this limit to be a finite, non-zero value, the power of $h$ must be zero. If $n-2 > 0$, the limit is $0$; if $n-2 < 0$, the limit is undefined.

Thus, $n-2 = 0$, which means **$n=2$**.

With $n=2$, the $h$ terms cancel out, leaving us with:
$$\frac{-2}{m} = -1$$

Solving this gives **$m=2$**. We have conquered the problem!

Select Answer:

Visualized Solution

Visualizing $∣ x - 1∣$

Finding $p$ (LHD)

Setting up the Limit

Substitution $x = 1 + h$

Logarithmic Power Rule

Standard Limit for Logarithm

Applying the Standard Limit

Standard Limit for Cosine

Simplifying the Expression

Matching Powers of $h$

Solving for $m$

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Geometry of the V-Shape

The Limit Setup

The Logarithmic and Trigonometric Dance

The Final Power Balance

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Visualized Solution

Visualizing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣x−1∣

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p (LHD)

Setting up the Limit

Substitution .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=1+h