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Math  /  Calculus

Question(b) lim⁡x→+∞x2+4x−x\quad \lim _{x \rightarrow+\infty} \sqrt{x^{2}+4 x}-x

Studdy Solution

STEP 1

1. We are dealing with a limit as x x approaches positive infinity.
2. The expression involves a square root and a linear term.
3. Simplification techniques such as rationalization or algebraic manipulation may be necessary.

STEP 2

1. Simplify the expression inside the limit.
2. Rationalize the expression to facilitate limit evaluation.
3. Evaluate the limit as x x approaches infinity.

STEP 3

First, observe the expression inside the limit:
lim⁡x→+∞x2+4x−x \lim _{x \rightarrow+\infty} \sqrt{x^{2}+4x}-x
To simplify, factor out x2 x^2 from the square root:
x2+4x=x2(1+4x) \sqrt{x^{2} + 4x} = \sqrt{x^{2}(1 + \frac{4}{x})}
This can be rewritten using the property of square roots:
=x1+4x = x\sqrt{1 + \frac{4}{x}}
Thus, the expression becomes:
lim⁡x→+∞x1+4x−x \lim _{x \rightarrow+\infty} x\sqrt{1 + \frac{4}{x}} - x

STEP 4

To rationalize and simplify further, factor out x x from the expression:
x(1+4x−1) x(\sqrt{1 + \frac{4}{x}} - 1)
Now, multiply and divide by the conjugate of the expression inside the parenthesis:
lim⁡x→+∞x(1+4x−1)(1+4x+1)1+4x+1 \lim _{x \rightarrow+\infty} \frac{x(\sqrt{1 + \frac{4}{x}} - 1)(\sqrt{1 + \frac{4}{x}} + 1)}{\sqrt{1 + \frac{4}{x}} + 1}
This simplifies to:
lim⁡x→+∞x((1+4x)−1)1+4x+1 \lim _{x \rightarrow+\infty} \frac{x((1 + \frac{4}{x}) - 1)}{\sqrt{1 + \frac{4}{x}} + 1}

STEP 5

Simplify the expression further:
lim⁡x→+∞x(4x)1+4x+1 \lim _{x \rightarrow+\infty} \frac{x(\frac{4}{x})}{\sqrt{1 + \frac{4}{x}} + 1}
This simplifies to:
lim⁡x→+∞41+4x+1 \lim _{x \rightarrow+\infty} \frac{4}{\sqrt{1 + \frac{4}{x}} + 1}

STEP 6

Evaluate the limit as x x approaches infinity. As x→∞ x \to \infty , 4x→0 \frac{4}{x} \to 0 , so:
1+4x→1=1 \sqrt{1 + \frac{4}{x}} \to \sqrt{1} = 1
Therefore, the limit becomes:
lim⁡x→+∞41+1=42=2 \lim _{x \rightarrow+\infty} \frac{4}{1 + 1} = \frac{4}{2} = 2
The value of the limit is:
2 \boxed{2}

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