Home
Class 12
MATHS
The value of "lt"(x to 1)[log(5) 5x]^(lo...

The value of `"lt"_(x to 1)[log_(5) 5x]^(log_(x) 5)` is:

A

1

B

`-1`

C

e

D

`e^(-1)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the limit \( \lim_{x \to 1} \left( \log_{5}(5x) \right)^{\log_{x}(5)} \), we can follow these steps: ### Step 1: Rewrite the expression We start by rewriting the limit expression: \[ \lim_{x \to 1} \left( \log_{5}(5x) \right)^{\log_{x}(5)} \] ### Step 2: Simplify the logarithm Using the property of logarithms, we can simplify \( \log_{5}(5x) \): \[ \log_{5}(5x) = \log_{5}(5) + \log_{5}(x) = 1 + \log_{5}(x) \] Thus, we can rewrite the limit as: \[ \lim_{x \to 1} \left( 1 + \log_{5}(x) \right)^{\log_{x}(5)} \] ### Step 3: Change the base of the logarithm Next, we can express \( \log_{x}(5) \) in terms of natural logarithms: \[ \log_{x}(5) = \frac{\log(5)}{\log(x)} \] Now our limit becomes: \[ \lim_{x \to 1} \left( 1 + \log_{5}(x) \right)^{\frac{\log(5)}{\log(x)}} \] ### Step 4: Analyze the limit As \( x \to 1 \), \( \log_{5}(x) \to 0 \) and \( \log(x) \to 0 \). This means we have an indeterminate form of the type \( 1^{\infty} \). To resolve this, we can take the natural logarithm of the expression: \[ y = \left( 1 + \log_{5}(x) \right)^{\frac{\log(5)}{\log(x)}} \] Taking the natural logarithm gives: \[ \ln(y) = \frac{\log(5)}{\log(x)} \cdot \ln(1 + \log_{5}(x)) \] ### Step 5: Use L'Hôpital's Rule Now, we need to evaluate: \[ \lim_{x \to 1} \frac{\log(5) \cdot \ln(1 + \log_{5}(x))}{\log(x)} \] This is again an indeterminate form \( \frac{0}{0} \). We can apply L'Hôpital's Rule: 1. Differentiate the numerator and denominator with respect to \( x \). 2. The derivative of \( \ln(1 + \log_{5}(x)) \) is \( \frac{1}{1 + \log_{5}(x)} \cdot \frac{1}{x \ln(5)} \). 3. The derivative of \( \log(x) \) is \( \frac{1}{x} \). ### Step 6: Evaluate the limit After applying L'Hôpital's Rule, we can evaluate the limit: \[ \lim_{x \to 1} \frac{\log(5) \cdot \frac{1}{1 + \log_{5}(x)} \cdot \frac{1}{x \ln(5)}}{\frac{1}{x}} = \lim_{x \to 1} \frac{\log(5)}{(1 + \log_{5}(x)) \ln(5)} \] As \( x \to 1 \), \( \log_{5}(x) \to 0 \), so this simplifies to: \[ \frac{\log(5)}{1 \cdot \ln(5)} = 1 \] ### Step 7: Exponentiate to find \( y \) Since \( \ln(y) \to 1 \), we have: \[ y \to e^{1} = e \] Thus, the final answer is: \[ \lim_{x \to 1} \left( \log_{5}(5x) \right)^{\log_{x}(5)} = e \]
Promotional Banner

Topper's Solved these Questions

  • LIMITS, CONTINUITY AND DIFFERENTIABILITY

    ML KHANNA|Exercise PROBLEM SET (1) (TRUE AND FALSE) |4 Videos
  • LIMITS, CONTINUITY AND DIFFERENTIABILITY

    ML KHANNA|Exercise PROBLEM SET (1) (FILL IN THE BLANKS) |7 Videos
  • INVERSE CIRCULAR FUNCTIONS

    ML KHANNA|Exercise Self Assessment Test|25 Videos
  • LINEAR PROGRAMMING

    ML KHANNA|Exercise Self Assessment Test|8 Videos

Similar Questions

Explore conceptually related problems

The value of lim_(x rarr1)(log_(5)5x)^(log_(x)5) is

The value of lim_(xrarr1)(log_5 5x)^(log_x5) , is

5^(log_(10)x)=50-x^(log_(10)5)

The number of value of x satisfying 1+log_(5)(x^(2)+1)>=log_(5)(x^(2)+4x+1) is

The value of x satisfying log_(3)4 -2 log_(3)sqrt(3x +1) =1 - log_(3)(5x -2)

If log_(5)[log_(3)(log_(2)x)]=1 , then x is:

If log_(5)[log_(3)(log_(2)x)]=1 then x is

x^(log_(5)x)>5 implies x in

Solve the value of x:2(log_(x)sqrt(5))^(2)-3log_(x)sqrt(5)+1=0