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How can I do question number in the following picture.

I assume you mean No 8.

This is going to be tricky, and I only have the first part for you. Perhaps you can use it to push the second part through.

Let y = (sin x/x)^1/x

ln y = 1/x[ln sin x - ln x] ==> If we find lim ln y, lim y = exp(lim ln y)

ln y = [ln sin x - ln x]/x

Diff(t&b):(l'Hospital's rule, to be repeated as many times as needed)

cos x/sin x - 1/x

-------------------

1

x cot x - 1

------------

x

Diff(t&b):(again)

x(-csc^2 x) + cot x

-x(1/ sin^2) + cos/sin

-x + sin cos

--------------

sin^2

Diff(t&b):

-1 + cos^2 - sin^2

------------------

2 sin cos

-1 + cos(2x)

--------------

sin(2x)

Diff(t&b):

-2 sin(2x)

-----------

2 cos(2x)

- sin(2x)

-----------

cos(2x)

As x -> 0, this -> 0/1 = 0.

So lim[x->0] ln y = 0

And so lim[x->0] y = 1.

Calculus

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All topics in first-year calculus including infinite series, max-min and related rate problems. Also trigonometry and complex numbers, theory of equations, exponential and logarithmic functions. I can also try (but not guarantee) to answer questions on Analysis -- sequences, limits, continuity.

I taught all mathematics subjects from elementary algebra to differential equations at a two-year college in New York City for 25 years.**Education/Credentials**

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