Question

Find the limit of each given transcendental function as x→c

a. lim x→-1 (2ln x+3/2x+5)

b. lim x→0 (e^3x - 1/ e^x -1)

c. lim x→0 (2^2x - 2^x+1 +1 / 2^x -1)

d. lim x→0 (3tan²x/2x²)

e. lim x→0 (1-cos2x/sin2x)

f. lim θ→0 (5θ/tan2θ)

g. lim x→0 (2x²-6x/sin2x)

h. lim t→0 [4t²/(1-cos² (t/2))]

Answer 1

a.

$x\xmapsto {lim}-1\space \frac{ln x+3}{2x+5}$

$=x\xmapsto{lim}-1\space \frac{ln x^2+3}{2x+5}$

$=\frac{ln(-1)^2+3}{2(-1)+5}$

$=\frac{0+3}{3}$

$=1$

b.

$x\xmapsto{lim}0\space \frac{e^{3x}-1}{e^x-1}$

$x\xmapsto{lim}0\space \frac{e^{3x}-1}{x}.\frac{1}{\frac{e^x-1}{x}}$

$=3(1).\frac{1}{1}=3$

c.

$x\xmapsto{lim}0\space \frac{2^{2x}-2^x+1}{2^x-1}$

$=\frac{1-1+1}{1-1}=\infin$

d.

According to L' Hospital's Rule

$x\xmapsto{lim}c\space \frac{f(x)}{g(x)}=x\xmapsto{lim}c\space \frac{f'(x)}{g'(x)}$

$x\xmapsto{lim}0\space \frac{3tan^2x}{2x^2}$

$\frac{3tan^2x}{2x^2}=\frac{0}{0}$ , so we can apply L' Hospital's rule

$x\xmapsto{lim}0\space \frac{3tan^2x}{2x^2}=x\xmapsto{lim}0\space \frac{\frac{d}{dx}3tan^2x}{\frac{d}{dx}2x^2}$

$=\frac{6}{4}[(1\times1)+0]=\frac{3}{2}$

e.

$x\xmapsto{lim}0\space \frac{1-cos2x}{sin2 x}=\frac{1-cos o}{sin 0}=\frac{1-1}{0}=0$

$=x\xmapsto{lim}0\space \frac{\frac{d}{dx}(1-cos2x}{\frac{d}{dx}(sin 2x)}$

$x\xmapsto{lim}0\frac{0-2\space sin 2 x}{2\space cos 2x}$

$=\frac{2 sin 0}{2 cos 0}=0$

f.

$\theta\xmapsto{lim}0\frac{5\theta}{tan2\theta}=0$

$\theta\xmapsto{lim}0\frac{5\theta}{tan 2\theta}=\theta\xmapsto{lim}0\space \frac{\frac{d}{dx}5\theta}{\frac{d}{dx}tan 2\theta}$

$=\frac{5}{2\space sin^22\theta}=\frac{5}{2\times 1}=\frac{5}{2}$

g.

$x\xmapsto{lim}0\frac{2x^2-6x}{sin 2x}$

$=x\xmapsto{lim}0\space \frac{\frac{d}{dx}(2x^2-6x)}{\frac{d}{dx}{sin 2x}}$

$x\xmapsto{lim}0\space \frac{2x^2-6x}{sin 2x}=-3$

h.

$x\xmapsto{lim}0\space \frac{4t^2}{1-cos^2 \frac{t}{2}}=x\xmapsto{lim}0\space \frac{4t^2}{2 sin^2t}$

$=x\xmapsto{lim}0\space \frac{8t}{4 \space sin \space t \space cos t}=x\xmapsto{lim}0\space \frac{8t}{2 \space sin \space 2t}$

$=\frac{8}{4 cos 0}=\frac{8}{4}=2$