Limit Evaluation: Step-by-Step Solution

We are asked to evaluate the following limit expression:

\[ \mathop {\lim }\limits_{x \to - 5} \frac{{\sqrt {\left( {2x + 35} \right)} - 5}}{{x + 5}} \]

Limit Form Analysis

First, let's substitute \(x = -5\) into the given limit expression to determine its initial form:

• For the numerator: \( \sqrt{2(-5) + 35} - 5 = \sqrt{-10 + 35} - 5 = \sqrt{25} - 5 = 5 - 5 = 0 \)
• For the denominator: \( -5 + 5 = 0 \)

Since both the numerator and the denominator evaluate to \(0\) when \(x = -5\), this limit is of the indeterminate form \( \frac{0}{0} \). This indicates that we need to use algebraic manipulation, such as rationalization, or calculus methods like L'Hopital's Rule to find the true value of the limit.

Limit Solution by Rationalization Method

The rationalization method is particularly useful when dealing with limits involving square roots that result in an indeterminate form. We achieve this by multiplying the numerator and denominator by the conjugate of the expression involving the square root.

The conjugate of the numerator \( \sqrt{\left( {2x + 35} \right)} - 5 \) is \( \sqrt{\left( {2x + 35} \right)} + 5 \).

Let \(L\) be the given limit:

\[ L = \mathop {\lim }\limits_{x \to - 5} \frac{{\sqrt {\left( {2x + 35} \right)} - 5}}{{x + 5}} \]

Multiply the numerator and denominator by the conjugate term:

\[ L = \mathop {\lim }\limits_{x \to - 5} \frac{{\sqrt {\left( {2x + 35} \right)} - 5}}{{x + 5}} \times \frac{{\sqrt {\left( {2x + 35} \right)} + 5}}{{\sqrt {\left( {2x + 35} \right)} + 5}} \]

Apply the difference of squares formula, \( (a-b)(a+b) = a^2 - b^2 \), to the numerator:

\[ L = \mathop {\lim }\limits_{x \to - 5} \frac{{\left( {\sqrt {\left( {2x + 35} \right)}} \right)^2 - 5^2}}{{\left( {x + 5} \right)\left( {\sqrt {\left( {2x + 35} \right)} + 5} \right)}} \]

Simplify the numerator:

\[ L = \mathop {\lim }\limits_{x \to - 5} \frac{{2x + 35 - 25}}{{\left( {x + 5} \right)\left( {\sqrt {\left( {2x + 35} \right)} + 5} \right)}} \]

\[ L = \mathop {\lim }\limits_{x \to - 5} \frac{{2x + 10}}{{\left( {x + 5} \right)\left( {\sqrt {\left( {2x + 35} \right)} + 5} \right)}} \]

Factor out \(2\) from the numerator:

\[ L = \mathop {\lim }\limits_{x \to - 5} \frac{{2\left( {x + 5} \right)}}{{\left( {x + 5} \right)\left( {\sqrt {\left( {2x + 35} \right)} + 5} \right)}} \]

Since \(x \to -5\), \(x \neq -5\), which means \(x + 5 \neq 0\). Therefore, we can cancel the common factor \( (x + 5) \) from both the numerator and the denominator:

\[ L = \mathop {\lim }\limits_{x \to - 5} \frac{{2}}{{\sqrt {\left( {2x + 35} \right)} + 5}} \]

Now, substitute \( x = -5 \) into the simplified expression:

\[ L = \frac{{2}}{{\sqrt {\left( {2(-5) + 35} \right)} + 5}} \]

\[ L = \frac{{2}}{{\sqrt {\left( {-10 + 35} \right)} + 5}} \]

\[ L = \frac{{2}}{{\sqrt {\left( {25} \right)} + 5}} \]

\[ L = \frac{{2}}{{5 + 5}} \]

\[ L = \frac{{2}}{{10}} \]

\[ L = \frac{{1}}{{5}} \]

Limit Calculation using L'Hopital's Rule

L'Hopital's Rule is an alternative method applicable when a limit is of an indeterminate form (\( \frac{0}{0} \) or \( \frac{\infty}{\infty} \)). It states that if \( \mathop {\lim }\limits_{x \to c} \frac{{f(x)}}{{g(x)}} \) is indeterminate, then \( \mathop {\lim }\limits_{x \to c} \frac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to c} \frac{{f'(x)}}{{g'(x)}} \), provided the latter limit exists.

Let \( f(x) = \sqrt{\left( {2x + 35} \right)} - 5 \) (the numerator) and \( g(x) = x + 5 \) (the denominator).

First, we find the derivatives of \( f(x) \) and \( g(x) \):

• Derivative of \( f(x) = \sqrt{\left( {2x + 35} \right)} - 5 \):

  Recall that \( \frac{d}{dx}\sqrt{u} = \frac{1}{2\sqrt{u}} \frac{du}{dx} \). Here, \( u = 2x + 35 \), so \( \frac{du}{dx} = 2 \).

  \[ f'(x) = \frac{1}{2\sqrt{\left( {2x + 35} \right)}} \times 2 - 0 = \frac{1}{\sqrt{\left( {2x + 35} \right)}} \]

• Derivative of \( g(x) = x + 5 \):

  \[ g'(x) = \frac{d}{dx}(x) + \frac{d}{dx}(5) = 1 + 0 = 1 \]

Now, apply L'Hopital's Rule by taking the limit of the ratio of the derivatives:

\[ L = \mathop {\lim }\limits_{x \to - 5} \frac{{f'(x)}}{{g'(x)}} = \mathop {\lim }\limits_{x \to - 5} \frac{{\frac{1}{{\sqrt {\left( {2x + 35} \right)}}}}}{{1}} \]

\[ L = \mathop {\lim }\limits_{x \to - 5} \frac{1}{{\sqrt {\left( {2x + 35} \right)}}} \]

Substitute \( x = -5 \) into the expression:

\[ L = \frac{1}{{\sqrt {\left( {2(-5) + 35} \right)}}} \]

\[ L = \frac{1}{{\sqrt {\left( {-10 + 35} \right)}}} \]

\[ L = \frac{1}{{\sqrt {\left( {25} \right)}}} \]

\[ L = \frac{1}{{5}} \]

Final Result of the Limit

Both the rationalization method and L'Hopital's Rule yield the same result. Therefore, the value of the limit is \( \frac{1}{5} \).