. Find the value of α such that limx→−1 2x2−αx−14x2−2x−3 exists. Find the limit.2. Let limx→0 f (x)x2 = 5. Show that limx→0 f (x)x = 0.3. Let f : R → R and x0 ∈ R. Suppose limx→x0 f (x) exists. Show that limx→0 f (x + x0) =limx→x0 f (x).4. Let f (x) = |x| for every x ∈ R. Show that f is continuous on R.5. Let f : [0, π] → R be defined by f (0) = 0 and f (x) = xsin 1x − 1x cos 1x for x̸ = 0. Is fcontinuous ?6. Let f, g : R → R be continuous such that given any two points x1 < x2, there exists apoint x3 such that x1 < x3 < x2 and f (x3) = g(x3). Show that f (x) = g(x) for all x.

To find the value of α such that limx→−1 2x2−αx−14x2−2x−3 exists, we need to evaluate the limit as x approaches -1.

Given limx→0 f (x)x2 = 5, we want to show that limx→0 f (x)x = 0.

Suppose f : R → R and x0 ∈ R, and limx→x0 f (x) exists. We need to prove that limx→0 f (x + x0) = limx→x0 f (x).

Let f (x) = |x| for every x ∈ R. We want to show that f is continuous on R.

For f : [0, π] → R, where f (0) = 0 and f (x) = xsin(1/x) − (1/x)cos(1/x) for x ≠ 0, we need to determine if f is continuous.

Given f, g : R → R are continuous functions such that for any two points x1 < x2, there exists a point x3 such that x1 < x3 < x2 and f (x3) = g(x3), we need to prove that f (x) = g(x) for all x.