Question

Evaluate $\lim_{x\to 1}\frac{x^2+x\log_e x-\log_ex-1}{(x^2-1)}.$

Answer 1

1

Answer 2

The given limit is: $\lim_{x\to 1}\frac{x^2+x\log_e x-\log_ex-1}{(x^2-1)}$

First, substitute $x=1$ into the expression to determine its form:

Numerator: $1^2 + 1 \cdot \log_e 1 - \log_e 1 - 1 = 1 + 1 \cdot 0 - 0 - 1 = 0$

Denominator: $1^2 - 1 = 0$

Since the limit is in the $\frac{0}{0}$ indeterminate form, we can use either L'Hopital's Rule or algebraic manipulation.

Method 1: Using L'Hopital's Rule

Let $f(x) = x^2+x\log_e x-\log_ex-1$ and $g(x) = x^2-1$. We need to find the derivatives of $f(x)$ and $g(x)$ with respect to $x$.

Derivative of the numerator $f'(x)$: $f'(x) = \frac{d}{dx}(x^2+x\log_e x-\log_ex-1)$ Using the product rule for $x\log_e x$: $\frac{d}{dx}(uv) = u'v + uv'$. Here $u=x, v=\log_e x$, so $u'=1, v'=\frac{1}{x}$. $f'(x) = 2x + (1 \cdot \log_e x + x \cdot \frac{1}{x}) - \frac{1}{x} - 0$ $f'(x) = 2x + \log_e x + 1 - \frac{1}{x}$

Derivative of the denominator $g'(x)$: $g'(x) = \frac{d}{dx}(x^2-1)$ $g'(x) = 2x$

Now, apply L'Hopital's Rule: $\lim_{x\to 1}\frac{f'(x)}{g'(x)} = \lim_{x\to 1}\frac{2x + \log_e x + 1 - \frac{1}{x}}{2x}$ Substitute $x=1$: $\frac{2(1) + \log_e 1 + 1 - \frac{1}{1}}{2(1)} = \frac{2 + 0 + 1 - 1}{2} = \frac{2}{2} = 1$

Method 2: Algebraic Manipulation

Factorize the numerator and the denominator.

Numerator: $x^2+x\log_e x-\log_ex-1$

Group terms: $(x^2-1) + (x\log_e x-\log_ex)$

Factor out common terms: $(x-1)(x+1) + \log_e x (x-1)$

Factor out $(x-1)$: $(x-1)[(x+1) + \log_e x]$

Denominator: $x^2-1$

Factorize: $(x-1)(x+1)$

Now, substitute these back into the limit expression: $\lim_{x\to 1}\frac{(x-1)[(x+1) + \log_e x]}{(x-1)(x+1)}$ Since $x \to 1$, $x \neq 1$, so we can cancel the common term $(x-1)$: $\lim_{x\to 1}\frac{(x+1) + \log_e x}{(x+1)}$ Substitute $x=1$: $\frac{(1+1) + \log_e 1}{(1+1)} = \frac{2 + 0}{2} = \frac{2}{2} = 1$

Both methods yield the same result.

Explanation of the solution:

The limit is in the $\frac{0}{0}$ indeterminate form.

Method 1 (L'Hopital's Rule): Differentiate the numerator and denominator separately. $f'(x) = 2x + \log_e x + 1 - \frac{1}{x}$ and $g'(x) = 2x$. Substitute $x=1$ into $\frac{f'(x)}{g'(x)}$ to get $\frac{2+0+1-1}{2} = \frac{2}{2} = 1$.

Method 2 (Algebraic Manipulation): Factor the numerator as $(x-1)(x+1) + \log_e x (x-1) = (x-1)[(x+1) + \log_e x]$. Factor the denominator as $(x-1)(x+1)$. Cancel the common $(x-1)$ term and substitute $x=1$ into the simplified expression $\frac{(x+1) + \log_e x}{(x+1)}$ to get $\frac{1+1+0}{1+1} = \frac{2}{2} = 1$.

The limit evaluates to 1.