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32.  If

f (x) = \left\{\begin{matrix} mx^2 + n & x < 0 \\ nx + m & 0 \leq x \leq 1 \\ nx^3+m & x > 1 \end{matrix}\righ t.
. For what integers m and n does both \lim_{x \rightarrow 0}f (x) \: \: and\: \: \lim_{x \rightarrow 1} f (x)  exist ? 

Given, Case 1: Limit at x = 0  The right-hand Limit or  Limit at  The left-hand limit or Limit at  Hence Limit will exist at x = 0 when m = n . Case 2: Limit at x = 1 The right-hand Limit or  Limit at  The left-hand limit or Limit at  Hence Limit at 1 exists at all integers.

30. If  f (x) = \left\{\begin{matrix} |x| + 1 & x < 0 \\ 0 & x = 0 \\ |x| -1& x > 0 \end{matrix}\right.  For what value (s) of a does \lim_{x \rightarrow a } f (x) exists ? 

Limit at x = a exists when the right-hand limit is equal to the left-hand limit. So, Case 1: when a = 0 The right-hand Limit or  Limit at  The left-hand limit or Limit at  Since Left-hand limit and right-hand limit are not equal, The limit of this function at x = 0 does not exists. Case 2: When a < 0  The right-hand Limit or  Limit at  The left-hand limit or Limit at  Since LHL = RHL,...

28.   Suppose 

f (x) = \left\{\begin{matrix} a+bx & x < 1 \\ 4 & x = 1 \\ b - ax & x > 1 \end{matrix}\right.   f (x) = f (1) what are possible values of a and b?

Given, And   Since the limit exists, left-hand limit = Right-hand limit = f(1). Left-hand limit  = f(1) Right-hand limit From both equations, we get that,  and  Hence the possible value of a and b are 0 and 4 respectively.

27.   Find \lim_{x \rightarrow 5 } f (x) , where f( x) = |x| -5

The right-hand Limit or  Limit at  The left-hand limit or Limit at  Since Left-hand limit and right-hand limit are equal, The limit of this function at x = 5 is 0.

26. Evaluate  \lim_{x \rightarrow 0} f (x) , \: \: where \: \: f (x) = \left\{\begin{matrix} \frac{x}{|x|} & x \neq 0 \\ 0 & x = 0 \end{matrix}\right.

The right-hand Limit or  Limit at  The left-hand limit or Limit at  Since Left-hand limit and right-hand limit are not equal, The limit of this function at x = 0 does not exists.

25. Evaluate \lim_{x \rightarrow 0} f (x) , \: \: where \: \: f (x) = \left\{\begin{matrix} \frac{|x|}{x} & x \neq 0 \\ 0 & x = 0 \end{matrix}\right.

The right-hand Limit or  Limit at  The left-hand limit or Limit at  Since Left-hand limit and right-hand limit are not equal, The limit of this function at x = 0 does not exists.

24. Find \lim_{x \rightarrow 1} f (x ) , \: \:where \: \: f (x) = \left\{\begin{matrix} x^2 -1 & x \neq 0 \\ -x^2 -1 & x > 1 \end{matrix}\right.

Limit at  Limit at  As we can see that Limit at  is not equal to Limit at ,The limit of this function at x = 1 does not exists.

23.   Find  \lim_{x \rightarrow 0} f (x ) \: \: \lim_{x \rightarrow 1} f (x) \: \: where \: \: \: f (x) = \left\{\begin{matrix} 2x+3 & x \leq 0 \\ 3 ( x+1)& x > 0 \end{matrix}\right.

Given Function Now, Limit at x = 0  :   : Hence limit at x = 0 is 3. Limit at x = 1 Hence limit at x = 1 is 6.    

22.   Evaluate the following limits \lim_{x \rightarrow \pi /2 } \frac{\tan 2x }{x - \pi /2 }

The function takes zero by zero form when the limit is put directly, so we simplify the function and then put the limits, So Let's put   Since we are changing the variable, limit will also change. as   So function in new variable becomes, As we know tha property       (Answer)

21.   Evaluate the following limits \lim_{x \rightarrow 0} \left ( \csc x - \cot x \right )

On putting the limit directly the function takes infinity by infinity form, So we simplify the function and then put the limit    (Answer)  

20.   Evaluate the following limits \lim_{x\rightarrow 0} \frac{\sin ax + bx }{ax + \sin bx } a,b ,a + b \neq 0

The function takes the zero by zero form when we put the limit into the function directly, so we try to eliminate this case by simplifying the function. So   (Answer)

19.  Evaluate the following limits \lim_{x \rightarrow 0} x \sec x

As function doesn't create any abnormality on putting the limit directly,we can put limit directly. So, .  (Answer)

18.   Evaluate the following limits \lim_{x \rightarrow 0}\frac{ax + x \cos x }{b \sin x }

The function takes the form zero by zero when we put the limit directly in the function,. since function consist of sin function and cos function, we try to make the function in the form of  as we know that it tends to 1 when x tends to 0. So,    (Answer)  

17.   Evaluate the following limits  \lim_{x\rightarrow 0} \frac{\cos 2x -1}{\cos x -1}

The limit:    The function takes the zero by zero form when the limit is put directly, so we simplify the function and then put the limit              (Answer)

16.  Evaluate the following limits \lim_{x\rightarrow 0}\frac{\cos x }{\pi -x }

The limit the function behaves well on directly putting the limit,so we put the limit directly. So.    (Answer)

15.   Evaluate the following limits \lim_{x \rightarrow \pi } \frac{\sin ( \pi -x )}{\pi ( \pi -x)}

The limit   (Answer)

14.   Evaluate the following limits  \lim_{x \rightarrow 0} \frac{\sin ax }{\sin bx } , a,b \neq 0

The limit, On putting the limit directly, the function takes the zero by zero  form  So,we convert it in the form of .and then put the limit,    (Answer)

13.   Evaluate the following limits \lim_{x \rightarrow 0 } \frac{\sin ax }{bx }

The limit Here on directly putting the limits, the function becomes  form. so we try to make the function in the form of . so, As      (Answer)
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