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  • Find  <img alt="\mathrm{\lim _{x \rightarrow 0} \frac{1+\sin x-\cos x}{1+\sin (p x)-\cos (p x)}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bx%20%5Crigh

Find  \mathrm{\lim _{x \rightarrow 0} \frac{1+\sin x-\cos x}{1+\sin (p x)-\cos (p x)}}

Answers (1)

best_answer

We are given the limit expression:

$$
\lim_{x \to 0} \frac{1 + \sin x - \cos x}{1 + \sin(px) - \cos(px)}
$$

Let’s solve this step-by-step using **Taylor expansion** (or small-angle approximation) around $x \to 0$:

---

### **Step 1: Taylor Expansion near 0**

$$
\sin x \approx x - \frac{x^3}{6}, \quad \cos x \approx 1 - \frac{x^2}{2}
$$

So the **numerator** becomes:

$$
1 + \sin x - \cos x \approx 1 + \left(x - \frac{x^3}{6}\right) - \left(1 - \frac{x^2}{2}\right)
= x - \frac{x^3}{6} + \frac{x^2}{2}
$$

Ignoring higher order terms ($x^3$ and smaller) near 0:

$$
\text{Numerator} \approx x + \frac{x^2}{2}
$$

Similarly, for the **denominator**:

$$
\sin(px) \approx px - \frac{(px)^3}{6}, \quad \cos(px) \approx 1 - \frac{(px)^2}{2}
$$

$$
1 + \sin(px) - \cos(px) \approx 1 + \left(px - \frac{(px)^3}{6} \right) - \left(1 - \frac{(px)^2}{2} \right)
= px + \frac{(px)^2}{2} - \frac{(px)^3}{6}
$$

Again, ignoring higher-order terms:

$$
\text{Denominator} \approx px + \frac{p^2x^2}{2}
$$

---

### **Step 2: Take the Limit**

So the limit becomes:

$$
\lim_{x \to 0} \frac{x + \frac{x^2}{2}}{px + \frac{p^2x^2}{2}}
$$

Divide numerator and denominator by $x$ (since $x \to 0$, $x \ne 0$):

$$
= \lim_{x \to 0} \frac{1 + \frac{x}{2}}{p + \frac{p^2x}{2}} \to \frac{1}{p}
$$

---

###Final Answer:

$$
\boxed{\frac{1}{p}}
$$

Posted by

Saumya Singh

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