1) [10] Compute the value of the following limits using...

Transcript

00:02 Hello, let's have a look at the question.

00:03 So here we have to solve the limit for the given questions.

00:07 Now in the a part we have limit x tends to 8 of x square minus 9x plus 8 and divided by x square minus 10x plus 16.

00:22 So now here we can write limit x tends to 8.

00:26 The numerator can be now written as x square minus 8.

00:30 X minus x plus 8 and the denominator can be written as x squared minus 8 x minus 2x and plus 60.

00:41 So now we can write limit x tends to 8 and numerator can be written as x minus 8 into x minus 1 and the denominator can be written as x minus 8 into x minus 2.

00:55 So now this will get cancelled here.

00:57 So now here we can write limit x tends to 8 and and x minus 1 divided by x minus 2.

01:05 So on putting the limit we will get 8 minus 1 divided by 8 minus 2, that is 7 by 6.

01:13 Now for the b part, we have limit x tends to infinity, x to the power 5 minus x to the power 4 plus 17 x to the power 3 divided by 3 x to the power 5 minus x square plus x and plus 6.

01:32 1.

01:33 So now here we can write limit x tends to infinity and here let us take x to the power 5 common.

01:41 So we have 1 minus 1 upon x plus 17 upon x square and divided by here also taking x to the power 5 common we have 3 minus 1 upon x to the power 3 plus 1 upon x to the power 4 plus 1 upon x to the power 5.

02:01 So now here we will have limit x tends to infinity and here we will get 1 minus 1 upon x plus 17 upon x squared divided by 3 minus 1 upon x to the power 3 plus 1 upon x to the power 4 and plus 1 upon x to the power 5...

Question

1) [10] Compute the value of the following limits using L'Hôpital's Rule: a) $\lim_{x \to 0} \frac{e^{4x} - 1 - 4x}{cos(2x) - 1} =$ b) $\lim_{x \to \infty} \frac{x^2}{e^{2x}} = $

Question

1) [10] Compute the value of the following limits using L'Hôpital's Rule: a) $\lim_{x \to 0} \frac{e^{4x} - 1 - 4x}{cos(2x) - 1} =$ b) $\lim_{x \to \infty} \frac{x^2}{e^{2x}} = $

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