00:01
So the first limit given to you as limit x tends to 2, 2x square minus 8 over 3x minus 6.
00:13
You can put the limit and you can observe that it is a 0 by 0 form.
00:18
So by l hospital rule, which states that if the numerator and denominator will appear to be in 0 by 0 form, then differentiate numerator and denominator if they are differentiable and then apply the limit.
00:41
So differentiating the numerator and the denominator, so limit x tends to 2, differentiating 2x square, it would be 4x over it is 3.
00:54
Now applying the limit, so it is equal to 4 times 2 over 3, which is equal to 8 by 3.
01:02
So the final answer is 8 by 3.
01:05
The second limit given to you as limit x tends to 7, x minus 6 over root 2x plus 2.
01:17
Applying the limit directly, so it is equal to 7 minus 6 over under root 2 times 7, it's 14 plus 2, which is equal to 1 over under root 16, which is equal to 1 over 4.
01:35
So the final answer is equal to 1 over 4.
01:39
The third limit which is given is limit x tends to infinity 2 minus 5x square over 7x square plus 8x plus 1.
01:56
Solving this equation or the limit is equal to x square common from the numerator, it is 2 over x square minus 5 upon x square from the denominator, it is 7 plus 8 over x plus 1 over x square.
02:15
Applying the limit, which is limit x tends to infinity x square 2 upon x square minus 5 upon x square over 7 plus 8 over x plus 1 over x square, x square and x square got cancelled...