Evaluate the following limits. If you use l'Hospital's Rule, be sure to indicate

Question

Evaluate the following limits. If you use lHospitals Rule, be sure to indicate when yyou are using it, and why it applies.
a)

lim_{x \to \infty} (3 \cdot 2^{1 - x} + x^{2} \cdot 2^{1 - x})

b)

lim_{x \to 0^{+}} {(1 + 5 x)}^{\frac{2}{x}}

Answer 1

Evaluate limits:

lim_{x \to \infty} (3 \cdot 2^{1 - x} + x^{2} \cdot 2^{1 - x})

lim_{x \to a} [f (x) \pm g (x)] = lim_{x \to a} f (x) \pm lim_{x \to a} g (x)

lim_{x \to \infty} (3 \cdot 2^{1 - x} + x^{2} \cdot 2^{1 - x}) = lim_{x \to \infty} (3 \cdot 2^{1 - x}) + lim_{x \to \infty} (x^{2} \cdot 2^{- x})

= 3 \cdot lim_{x \to \infty} (2^{1 - x}) + 2 \cdot lim_{x \to \infty} (x^{2} \frac{1}{2^{x}})

Take

3 \cdot lim_{x \to \infty} (2^{1 - x})

Apply exponent rule

a^{x} = e^{\ln (a^{x})} = e^{x \ln (a)}

= 3 \cdot lim_{x \to \infty} (e^{(1 - x) \ln (2)}))

= 3 \cdot lim_{x \to \infty} (2^{1 - x})

= 3 \cdot lim_{x \to - \infty} (e^{x \ln (2)})

Apply the Limit Chain Rule:

g (x) = x \ln (2), f (u) = e^{u}

= - \infty \cdot \ln (2)

= - \infty

3 \cdot lim_{x \to - \infty} (e^{x \ln (2)}) = 3 \cdot e^{- \infty}

= 3 \cdot 0

= 0

Take

2 \cdot lim_{x \to \infty} (x^{2} \frac{1}{2^{x}})

If

\sum a_{n}

converges, then

lim_{n \to \infty} (a_{n}) = 0

Apply ratio test and check weather series is convergent or divergent.
If

| \frac{a_{n + 1}}{a_{n}} | \leq q

eventually for some 0<q<1, the

Answer 2

b) Evaluate limits:

lim_{x \to 0^{+}} {(1 + 5 x)}^{\frac{2}{x}}

Apply exponent rule:

a^{x} = e^{\ln (a^{x})} = e^{x \cdot \ln (a)}

{(1 + 5 x)}^{\frac{2}{x}} = e^{\frac{2}{x} \ln (1 + 5 x)}

Apply the limit Chain Rule:

g (x) = \frac{2}{x} \ln (1 + 5 x), f (u) = e^{u}

lim_{x \to 0^{+}} g (x) = lim_{x \to 0^{+}} \frac{2}{x} \ln (1 + 5 x)

= 2 \cdot lim_{x \to 0^{+}} (\frac{\ln (1 + 5 x)}{x})

lim_{x \to 0^{+}} (\frac{\ln (1 + 5 x)}{x}) = \frac{0}{0}

LHopital

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