JEE Mains Model Paper 8

We have

lim

x→0−

f (x) =

lim

h→0+

(sin (-h) + cos (−h)cosec(−h)
=

lim

h→0+

(cosh−sinh)−cosech

lim

h→0+

(1 + (cosh - sinh - 1))

1

(cosh−sinh−1)

.

(cosh−sinh−1)

(−sinh)

=

lim

h→0+

e

cosh−sinh−1

−sinh

= e
=

lim

h→0+

e

cosh−sinh−1

−sinh

= e
Now we have

lim

x→0+

f (x) =

lim

h→0+

e1∕h+e2∕h+e3∕h

ae−2+1∕h+be−1+3h

=

lim

h→0+

e2∕h+e−1∕h+1

(ae−2e−2∕h+(be−1)

=

e

b

If 'f' is continuous at x = 0, then e = a =

e

b

gives a = e and b = 1