Evaluate the following limit. If you use L'Hospital Rule, be sure to indicate when you are using it, and why it applies. lim_{xrightarrowinfty}(3cdot2^{1-x}+x^2cdot2^{1-x})

Question

Evaluate the following limit. If you use LHospital

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^{1 - x})

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

\sum_{n = 1}^{\infty} | a_{n} |

converges.
If

| \frac{a_{n + 1}}{a_{n}} | > 1

eventually then

\sum_{n = 1}^{\infty} a_{n}

diverges.

lim_{x \to \infty} (\frac{x^{2}}{2^{x}}) = 0

= 2 \cdot 0

= 0

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

= 0 + 0

= 0

Result:

lim_{x \to \infty} (3 \cdot 2^{1 - x} + x^{2} \cdot 2^{1 - x}) = 0

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