Key to "Evaluate the following limits. a) limx→38+20x3−8x b) limx→38−20x3−8x"

Tomasettiq3

Answered question

2021-11-16

Evaluate the following limits.
a)

lim_{x \to {\frac{3}{8}}^{+}} \frac{20 x}{3 - 8 x}

lim_{x \to {\frac{3}{8}}^{-}} \frac{20 x}{3 - 8 x}

tnie54

Beginner2021-11-17Added 18 answers

The given limits are:
a)

lim_{x \to {\frac{3}{8}}^{+}} \frac{20 x}{3 - 8 x}

as we know that

lim_{x \to a^{+}} f (x) = lim_{h \to 0} f (a + h)

therefore,

lim_{x \to {\frac{3}{8}}^{+}} \frac{20 x}{3 - 8 x} = lim_{h \to 0} \frac{20 (\frac{3}{8} + h)}{3 - 8 (\frac{3}{8} + h)}

= lim_{h \to 0} \frac{20 (\frac{3}{8} + h)}{3 - 8 \times \frac{3}{8} - 8 h}

= lim_{h \to 0} \frac{20 (\frac{3}{8} + h)}{3 - 3 - 8 h}

= lim_{h \to 0} \frac{20 (\frac{3}{8} + h)}{- 8 h}

= lim_{h \to 0} \frac{20 (\frac{3}{8} + h)}{h}

= - \frac{1}{8} lim_{h \to 0} \frac{20 (\frac{3}{8} + h)}{h}

= - \frac{1}{8} \times \frac{20 (\frac{3}{8} + 0)}{0}

= - \frac{1}{8} \times \frac{\frac{15}{2}}{0}

= - \frac{1}{8} \times \infty

= - \infty

James Obrien

Beginner2021-11-18Added 16 answers

lim_{x \to {\frac{3}{8}}^{-}} \frac{20 x}{3 - 8 x}

as we know that

lim_{x \to a^{-}} f (x) = lim_{h \to 0} f (a - h)

therefore,

lim_{x \to {\frac{3}{8}}^{-}} \frac{20 x}{3 - 8 x} = lim_{h \to 0} \frac{20 (\frac{3}{8} - h)}{3 - 8 (\frac{3}{8} - h)}

= lim_{h \to 0} \frac{20 (\frac{3}{8} - h)}{3 - 8 \times \frac{3}{8} + 8 h}

= lim_{h \to 0} \frac{20 (\frac{3}{8} - h)}{3 - 3 + 8 h}

= lim_{h \to 0} \frac{20 (\frac{3}{8} - h)}{8 h}

= lim_{h \to 0} \frac{20 (\frac{3}{8} - h)}{h}

= \frac{1}{8} lim_{h \to 0} \frac{20 (\frac{3}{8} - h)}{h}

= \frac{1}{8} \times \frac{20 (\frac{3}{8} - 0)}{0}

= \frac{1}{8} \times \frac{\frac{15}{2}}{0}

= \frac{1}{8} \times \infty

= \infty

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?