Question

2. Value of limx - 0(1+Sin(x))Cosec(x)​

Answer 1

Answer:

Let the limit = L

L = lim x-> 0 ( 1 + sinx )^cosecx

logL = lim x-> 0 cosecx [ log( 1 + sinx ) ]

logL = lim x-> 0 [ log( 1 + sinx ) ] / sinx

This is a 0/0 form.

Hence we apply the L'hospital rule.

logL = lim x-> 0 [ 1/( 1 + sinx ) ] cosx/cosx

logL = 1

L = e.

You can also use the series expression for log(1+x).

There is a general way to evaluate limits where after direct substitution, you get something like 1^(infinity)

If L = lim x -> a f(x)^[g(x)]

And f(a) = 1 and lim x-> a g(x) = infinity,

Then,

logL = lim x -> a [ f(x) - 1 ] g(x)

L = e^{ lim x -> a [ f(x) - 1 ] g(x) }

Step-by-step explanation:

hope this helps you dear

and have a great day

Answer 2

Answer:

Let the limit = L

L = lim x-> 0 ( 1 + sinx )^cosecx

logL = lim x-> 0 cosecx [ log( 1 + sinx ) ]

logL = lim x-> 0 [ log( 1 + sinx ) ] / sinx

This is a 0/0 form.

Hence we apply the L'hospital rule.

logL = lim x-> 0 [ 1/( 1 + sinx ) ] cosx/cosx

logL = 1

L = e.

You can also use the series expression for log(1+x).

There is a general way to evaluate limits where after direct substitution, you get something like 1^(infinity)

If L = lim x -> a f(x)^[g(x)]

And f(a) = 1 and lim x-> a g(x) = infinity,

Then,

logL = lim x -> a [ f(x) - 1 ] g(x)

L = e^{ lim x -> a [ f(x) - 1 ] g(x) }