Questions: Calculer lorsqu'elles existent les limites suivantes a) lim x → 0 (x^2+2x)/x b) lim x → -∞ (x^2+2x)/x c) lim x → 2 (x^2-4)/(x^2-3x+2) d) lim x → π (sin^2 x)/(1+cos x) e) lim x → 0 (√(1+x)-√(1+x^2))/x f) lim x → +∞ √(x+5)-√(x-3) g) lim x → 0 (√[3](1+x^2)-1)/x^2 h) lim x → 1 (x-1)/(x^n-1)

Calculer lorsqu'elles existent les limites suivantes
a) lim x → 0 (x^2+2x)/x
b) lim x → -∞ (x^2+2x)/x
c) lim x → 2 (x^2-4)/(x^2-3x+2)
d) lim x → π (sin^2 x)/(1+cos x)
e) lim x → 0 (√(1+x)-√(1+x^2))/x
f) lim x → +∞ √(x+5)-√(x-3)
g) lim x → 0 (√[3](1+x^2)-1)/x^2
h) lim x → 1 (x-1)/(x^n-1)

Transcript text: Calculer lorsqu'elles existent les limites suivantes a) $\lim _{x \rightarrow 0} \frac{x^{2}+2|x|}{x}$ b) $\lim _{x \rightarrow-\infty} \frac{x^{2}+2|x|}{x}$ c) $\lim _{x \rightarrow 2} \frac{x^{2}-4}{x^{2}-3 x+2}$ d) $\lim _{x \rightarrow \pi} \frac{\sin ^{2} x}{1+\cos x}$ e) $\lim _{x \rightarrow 0} \frac{\sqrt{1+x}-\sqrt{1+x^{2}}}{x}$ f) $\lim _{x \rightarrow+\infty} \sqrt{x+5}-\sqrt{x-3}$ g) $\lim _{x \rightarrow 0} \frac{\sqrt[3]{1+x^{2}}-1}{x^{2}}$ h) $\lim _{x \rightarrow 1} \frac{x-1}{x^{n}-1}$

Solution

Solution Steps

Step 1: Limit as $ x $ Approaches 0

To evaluate the limit $ \lim_{x \rightarrow 0} \frac{x^2 + 2|x|}{x} $, we analyze the expression. As $ x $ approaches 0, the absolute value function behaves as follows:

For $ x \geq 0 $, $ |x| = x $, so the expression simplifies to $ \frac{x^2 + 2x}{x} = x + 2 $, which approaches $ 2 $.
For $ x < 0 $, $ |x| = -x $, so the expression simplifies to $ \frac{x^2 - 2x}{x} = x - 2 $, which approaches $ -2 $.

Since the left-hand limit and right-hand limit do not match, the limit does not exist. However, the right-hand limit is $ 2 $.

Step 2: Limit as $ x $ Approaches Negative Infinity

To evaluate the limit $ \lim_{x \rightarrow -\infty} \frac{x^2 + 2|x|}{x} $, we note that as $ x $ approaches negative infinity, $ |x| = -x $. Thus, the expression simplifies to: \[ \frac{x^2 + 2(-x)}{x} = \frac{x^2 - 2x}{x} = x - 2 \] As $ x $ approaches negative infinity, this expression approaches $ -\infty $.

Step 3: Limit as $ x $ Approaches 2

To evaluate the limit $ \lim_{x \rightarrow 2} \frac{x^2 - 4}{x^2 - 3x + 2} $, we factor both the numerator and the denominator:

The numerator $ x^2 - 4 $ factors to $ (x - 2)(x + 2) $.
The denominator $ x^2 - 3x + 2 $ factors to $ (x - 1)(x - 2) $.

Thus, the expression simplifies to: \[ \frac{(x - 2)(x + 2)}{(x - 1)(x - 2)} \] We can cancel $ (x - 2) $ (noting that $ x \neq 2 $ in the limit), leading to: \[ \frac{x + 2}{x - 1} \] Now, substituting $ x = 2 $: \[ \frac{2 + 2}{2 - 1} = \frac{4}{1} = 4 \]