find cos 0 el - (1 + h) 43. lim 0-0e" - 0 - 1 44. lim

Q: 2h 4. lim h0 2h S(x) dr

A: Answer and explanation is given below

Q: lim z= 10 2 0 -2 0 -10

A: z=x +i y we need to find  Lim z→i z^2

Q: 6h lim-0 tan4h

A:

Q: lim z2= 10 20 -2 0 -10

A: The objective is to find the limz→iz2. Since z2 is continuous.

Q: 111. lim cos x -2

A: To find,

Q: tan(23z) lim

A: This problem of limit can be solved by the L' Hospital rule which is explained in step 2.

Q: 3 Lim x40 Cosx (2 Cos²x - 1) Cosx -

A:

Q: whatin Lim (19x) tan(3x メ→0 2.

A: The limit is limx→0+9x tan3x+π2. Factor out the constant term.  limx→ 0+9xtan3x+π2=9·limx→…

Q: 1 2. Lim x0x 1 sin x

A: Given:      limx→0+1x-1sinx simplifying we get     =limx→0+sinx-xxsinx Apply L'Hospital's Rule:…

Q: e2x - e 2x 3. lim, 0 sinx

A: To evaluate the given limit.limx→0 e2x-e-2xsin x

Q: 61. lim n sin = T n-00

A:

Q: x + sin x 5. lim x 2x + 5

A: We have to solve:      limx→∞x+sin x2x+5

Q: Find lim Sin (X+2) X->2 x4x-2

A: Explained below

Q: x²+x-2 1. lim X-o 4x-1

A:

Q: 4 is x→00 2x2-3 lim A. undefined В. 4 С. 2 D. 0 E. 1

A: To compute the following limit

Q: 9. Simplify sec² x - 1 sec² x

A: secx=1cosxI=sec2x-1sec2xsin2x+cos2x=1

Q: lim,-0 e" = ?

A: Consider the given limit limx→-∞ex.

Q: 1 I 4 x 4 x 4 49. lim-

A:

Q: COS X lim x0x +3

A:

Q: 2x x-0 sin 1x f. lim

A:

Q: (1 + +)" =- 65. lim

A: To show: limn→∞1+1nn=e

Q: O Lim x+4 2. ズラー1 O Lim VE -1 x-1 O Lim X-

A:

Q: -1 Evaluate lim 3x А. 2/3 C. 1/3 В.-2 D. -3

A:

Q: sin 2x 2. lim- X0 2x+x

A:

Q: lim sin 1- 1

A: Given limit, limx→-∞sin1x

Q: 1 lim 2 -1 x* -1,

A:

Q: lim, 0 6 tan² (5x) 15x²

A:

Q: [[+] [[P] [smal

A:

Q: 4+ sin x d. lim - 1- i cos(x+x) +2 g. lim

A: To find out the limit, substitute 0 for x and find out the limit value

Q: Find y lnxy tan (X+9)

A:

Q: 5x² - 3x b. lim 7x² +1 sint? d. lim sin 5x f. lim 0-x

A:

Q: Find Lim 5e*-2x x-0 A. 5e C. 5 D. 0 B.

A: Solution

Q: C. lim (4x xx-√16x² + 3x - 5) X→∞

A: Given limits, (c) limx→∞4x-16x2+3x-5 (d) limx→03x-7x2x-15x We have to find the Limit of the given…