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Sagot :
To evaluate the limit
[tex]\[ \lim_{x \to -2^+} \left( \frac{1}{x+2} - \frac{1}{\sqrt{x+2}} \right), \][/tex]
we need to analyze the behavior of each term separately and see if we can simplify the expression.
First, let's identify the forms of the terms as [tex]\( x \)[/tex] approaches [tex]\(-2\)[/tex] from the right ([tex]\(x \rightarrow -2^+\)[/tex]):
1. [tex]\( \frac{1}{x+2} \)[/tex] becomes very large because [tex]\( x+2 \)[/tex] approaches [tex]\( 0^+ \)[/tex], making the term [tex]\( \frac{1}{x+2} \)[/tex] tend to [tex]\( +\infty \)[/tex].
2. [tex]\( \frac{1}{\sqrt{x+2}} \)[/tex] similarly becomes very large because [tex]\( \sqrt{x+2} \)[/tex] also approaches [tex]\( 0^+ \)[/tex], making the term [tex]\( \frac{1}{\sqrt{x+2}} \)[/tex] tend to [tex]\( +\infty \)[/tex].
Given that direct substitution leads to the indeterminate form [tex]\( \infty - \infty \)[/tex], we need to manipulate the expression to apply l'Hôpital's Rule.
Let's rewrite the expression inside the limit by finding a common denominator:
[tex]\[ \frac{1}{x+2} - \frac{1}{\sqrt{x+2}} = \frac{\sqrt{x+2} - (x+2)}{(x+2)\sqrt{x+2}}. \][/tex]
Now, simplify the numerator:
[tex]\[ \sqrt{x+2} - (x+2). \][/tex]
We have:
[tex]\[ \lim_{x \to -2^+} \frac{\sqrt{x+2} - (x+2)}{(x+2)\sqrt{x+2}}. \][/tex]
To handle this limit, let's use l'Hôpital's Rule, which requires taking derivatives of the numerator and denominator since the limit is still of the form [tex]\(\frac{0}{0}\)[/tex] as [tex]\( x \to -2^+ \)[/tex].
Calculate the derivative of the numerator and the denominator:
1. The derivative of the numerator:
[tex]\[ \frac{d}{dx} [\sqrt{x+2} - (x+2)] = \frac{1}{2\sqrt{x+2}} - 1. \][/tex]
2. The derivative of the denominator:
[tex]\[ \frac{d}{dx} [(x+2)\sqrt{x+2}] = \frac{d}{dx} [(x+2)^{3/2}] = \frac{3}{2} (x+2)^{1/2}. \][/tex]
Therefore, applying l'Hôpital's Rule, the limit becomes:
[tex]\[ \lim_{x \to -2^+} \frac{\frac{1}{2\sqrt{x+2}} - 1}{\frac{3}{2} \sqrt{x+2}}. \][/tex]
Simplify the fraction:
[tex]\[ = \lim_{x \to -2^+} \frac{1 - 2\sqrt{x+2}}{3(x+2)}. \][/tex]
As [tex]\( x \to -2^+ \)[/tex], the term [tex]\( 2\sqrt{x+2} \)[/tex] approaches [tex]\( 0 \)[/tex] much faster than [tex]\( 1 \)[/tex], so the fraction simplifies:
[tex]\[ \lim_{x \to -2^+} \frac{1 - 2\sqrt{x+2}}{3(x+2)} = \lim_{x \to -2^+} \frac{1 - 0}{3(x+2)} = \lim_{x \to -2^+} \frac{1}{3(x+2)}. \][/tex]
Since [tex]\(3(x + 2)\)[/tex] approaches 0 from the positive side:
[tex]\[ = \frac{1}{3} \times \infty = \infty. \][/tex]
So, the given limit evaluates to [tex]\( \infty \)[/tex].
[tex]\[ \lim_{x \to -2^+} \left( \frac{1}{x+2} - \frac{1}{\sqrt{x+2}} \right), \][/tex]
we need to analyze the behavior of each term separately and see if we can simplify the expression.
First, let's identify the forms of the terms as [tex]\( x \)[/tex] approaches [tex]\(-2\)[/tex] from the right ([tex]\(x \rightarrow -2^+\)[/tex]):
1. [tex]\( \frac{1}{x+2} \)[/tex] becomes very large because [tex]\( x+2 \)[/tex] approaches [tex]\( 0^+ \)[/tex], making the term [tex]\( \frac{1}{x+2} \)[/tex] tend to [tex]\( +\infty \)[/tex].
2. [tex]\( \frac{1}{\sqrt{x+2}} \)[/tex] similarly becomes very large because [tex]\( \sqrt{x+2} \)[/tex] also approaches [tex]\( 0^+ \)[/tex], making the term [tex]\( \frac{1}{\sqrt{x+2}} \)[/tex] tend to [tex]\( +\infty \)[/tex].
Given that direct substitution leads to the indeterminate form [tex]\( \infty - \infty \)[/tex], we need to manipulate the expression to apply l'Hôpital's Rule.
Let's rewrite the expression inside the limit by finding a common denominator:
[tex]\[ \frac{1}{x+2} - \frac{1}{\sqrt{x+2}} = \frac{\sqrt{x+2} - (x+2)}{(x+2)\sqrt{x+2}}. \][/tex]
Now, simplify the numerator:
[tex]\[ \sqrt{x+2} - (x+2). \][/tex]
We have:
[tex]\[ \lim_{x \to -2^+} \frac{\sqrt{x+2} - (x+2)}{(x+2)\sqrt{x+2}}. \][/tex]
To handle this limit, let's use l'Hôpital's Rule, which requires taking derivatives of the numerator and denominator since the limit is still of the form [tex]\(\frac{0}{0}\)[/tex] as [tex]\( x \to -2^+ \)[/tex].
Calculate the derivative of the numerator and the denominator:
1. The derivative of the numerator:
[tex]\[ \frac{d}{dx} [\sqrt{x+2} - (x+2)] = \frac{1}{2\sqrt{x+2}} - 1. \][/tex]
2. The derivative of the denominator:
[tex]\[ \frac{d}{dx} [(x+2)\sqrt{x+2}] = \frac{d}{dx} [(x+2)^{3/2}] = \frac{3}{2} (x+2)^{1/2}. \][/tex]
Therefore, applying l'Hôpital's Rule, the limit becomes:
[tex]\[ \lim_{x \to -2^+} \frac{\frac{1}{2\sqrt{x+2}} - 1}{\frac{3}{2} \sqrt{x+2}}. \][/tex]
Simplify the fraction:
[tex]\[ = \lim_{x \to -2^+} \frac{1 - 2\sqrt{x+2}}{3(x+2)}. \][/tex]
As [tex]\( x \to -2^+ \)[/tex], the term [tex]\( 2\sqrt{x+2} \)[/tex] approaches [tex]\( 0 \)[/tex] much faster than [tex]\( 1 \)[/tex], so the fraction simplifies:
[tex]\[ \lim_{x \to -2^+} \frac{1 - 2\sqrt{x+2}}{3(x+2)} = \lim_{x \to -2^+} \frac{1 - 0}{3(x+2)} = \lim_{x \to -2^+} \frac{1}{3(x+2)}. \][/tex]
Since [tex]\(3(x + 2)\)[/tex] approaches 0 from the positive side:
[tex]\[ = \frac{1}{3} \times \infty = \infty. \][/tex]
So, the given limit evaluates to [tex]\( \infty \)[/tex].
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