 Hello and how are you all today? My name is Priyanka and I shall be helping you with the following question. Let's see. Evaluate the following limits. Now here the limit which is given to us is limit x approaches 0 cos x minus cot x. Now here we need to find out the value of this limit. Now first of all we will be converting cos x and cot x into sin x and cos x. So we have limit x approaches 0 cos x can be written as 1 over sin x whereas cot x can be written as cos x by sin x right. Further we have limit x approaches 0 taking sin x common we have in the numerator 1 minus cos x right. Further we have limit x approaches 0. Now on rationalizing the numerator we have 1 minus cos x over sin x and multiplying the numerator and denominator with 1 plus cos x we have 1 minus cos square x divided by sin square x sorry divided by sin x getting multiplied by 1 plus cos x also. We know that sin square x plus cos square x is equal to 1 isn't it? So we have limit x approaches 0 1 minus cos square x can be written as exactly sin square x. So we have sin square x divided by sin x multiplied by 1 plus cos x. Further limit x approaches 0 we have sin x over 1 plus cos x. Now since here we cannot since we cannot use the trigonometric limits. So at x is equal to 0 we have the value of sin x over 1 plus cos x as sin 0 over 1 plus cos 0 that is further equal to 0 upon 1 plus 1 as the value of cos 0 is 1. So we have 0 over 2 that is further equal to 0. So the answer to the solution is 0 that is the value of limit x approaches to 0 cos x minus cot x right. This ends my solution hope you understood and enjoyed the session. Have a nice day ahead.