Core Logic
Let the limit be L$L$. First, separate the logarithm of the product into a sum of logarithms:
L = x arrow 0 ln( (ex)) + ln( (e² x)) + + ln( (e¹⁰ x))e² - e2 x$$L = \lim_{x \rightarrow 0} \frac{\ln(\sec(ex)) + \ln(\sec(e^2 x)) + \dots + \ln(\sec(e^{10} x))}{e^2 - e^{2 \cos x}}$$
Factor out e2 x$e^{2 \cos x}$ in the denominator:
e² - e2 x = e2 x ( e2 - 2 x - 1 )$$e^2 - e^{2 \cos x} = e^{2 \cos x} \left( e^{2 - 2 \cos x} - 1 \right)$$
Use the standard limit (e^t - 1)/(t) arrow 1$\frac{e^t - 1}{t} \rightarrow 1$ as t arrow 0$t \rightarrow 0$. Here, t = 2 - 2 x$t = 2 - 2 \cos x$.
Multiply and divide the denominator by (2 - 2 x)$(2 - 2 \cos x)$:
e2 x ( e2 - 2 x - 12 - 2 x ) (2 - 2 x)$$e^{2 \cos x} \left( \frac{e^{2 - 2 \cos x} - 1}{2 - 2 \cos x} \right) (2 - 2 \cos x)$$
As x arrow 0$x \rightarrow 0$, e2 x arrow e²$e^{2 \cos x} \rightarrow e^2$ and the bracket term arrow 1$\rightarrow 1$.
Furthermore, 2 - 2 x = 2(1 - x) ≈ 2 ((x²)/(2)) = x²$2 - 2 \cos x = 2(1 - \cos x) \approx 2 \left(\frac{x^2}{2}\right) = x^2$.
Step 1: Simplify Denominator
The denominator effectively behaves as e² · x²$e^2 \cdot x^2$ as x arrow 0$x \rightarrow 0$.
L = x arrow 0 ln( (ex)) + ln( (e² x)) + + ln( (e¹⁰ x))e² x²$$L = \lim_{x \rightarrow 0} \frac{\ln(\sec(ex)) + \ln(\sec(e^2 x)) + \dots + \ln(\sec(e^{10} x))}{e^2 x^2}$$
Step 2: Apply L'Hôpital's Rule
Since this is a (0)/(0)$\frac{0}{0}$ form, we apply L'Hôpital's rule by differentiating numerator and denominator with respect to x$x$:
Derivative of numerator: (d)/(dx) ln( (cx)) = (1)/( (cx)) · (cx) (cx) · c = c (cx)$\frac{d}{dx} \ln(\sec(cx)) = \frac{1}{\sec(cx)} \cdot \sec(cx)\tan(cx) \cdot c = c \tan(cx)$.
So the numerator derivative is e (ex) + e² (e² x) + + e¹⁰ (e¹⁰ x)$e \tan(ex) + e^2 \tan(e^2 x) + \dots + e^{10} \tan(e^{10} x)$.
Derivative of denominator: 2e² x$2e^2 x$.
L = x arrow 0 e (ex) + e² (e² x) + + e¹⁰ (e¹⁰ x)2e² x$$L = \lim_{x \rightarrow 0} \frac{e \tan(ex) + e^2 \tan(e^2 x) + \dots + e^{10} \tan(e^{10} x)}{2e^2 x}$$
Step 3: Evaluate Remaining Limit
Apply the standard limit ( (kx))/(x) = k$\frac{\tan(kx)}{x} = k$:
L = (1)/(2e²) ( e(e) + e²(e²) + + e¹⁰(e¹⁰) )$$L = \frac{1}{2e^2} \left( e(e) + e^2(e^2) + \dots + e^{10}(e^{10}) \right)$$
L = (1)/(2e²) ( e² + e⁴ + e⁶ + + e²⁰ )$$L = \frac{1}{2e^2} \left( e^2 + e^4 + e^6 + \dots + e^{20} \right)$$
This is a Geometric Progression with 10 terms, first term a = e²$a = e^2$, common ratio r = e²$r = e^2$.
Sum = a r¹⁰ - 1r - 1 = e² (e²)¹⁰ - 1e² - 1 = e²(e²⁰ - 1)e² - 1$$Sum = a \frac{r^{10} - 1}{r - 1} = e^2 \frac{(e^2)^{10} - 1}{e^2 - 1} = \frac{e^2(e^{20} - 1)}{e^2 - 1}$$
L = (1)/(2e²) · e²(e²⁰ - 1)e² - 1 = e²⁰ - 12(e² - 1)$$L = \frac{1}{2e^2} \cdot \frac{e^2(e^{20} - 1)}{e^2 - 1} = \frac{e^{20} - 1}{2(e^2 - 1)}$$
Chapter Mix
Class 11 Mathematics: Limits, Continuity and Differentiability
Class 11 Mathematics: Sequences and Series