Use l’hospital’s rule to evaluate the following limit.

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Step-by-step explanation:

lim(t→∞) [t ln(1 + 3/t) ]

If we evaluate the limit, we get:

∞ ln(1 + 3/∞)

∞ ln(1 + 0)

∞ 0

This is undetermined.  To apply L'Hopital's rule, we need to rewrite this so the limit evaluates to ∞/∞ or 0/0.

lim(t→∞) [t ln(1 + 3/t) ]

lim(t→∞) [ln(1 + 3/t) / (1/t)]

This evaluates to 0/0.  We can simplify a little with u substitution:

lim(u→0) [ln(1 + 3u) / u]

Applying L'Hopital's rule:

lim(u→0) [1/(1 + 3u) × 3 / 1]

lim(u→0) [3 / (1 + 3u)]

3 / (1 + 0)

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answer:

c. the third graph

because they could drop even lower..