Answers

2015-03-02T16:37:16+05:30

This Is a Certified Answer

×
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
We use the definition of e^x to solve this problem.

e^x =  \lim_{n \to \infty} (1+\frac{x}{n})^n\\\\= 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....+\frac{x^n}{n!}+....\infty\\\\e^x=1 + x(1+\frac{x}{2!}+\frac{x^2}{3!}+....+\frac{x^{n-1}}{n!}+....\infty)\\\\1)\\ \lim_{h \to 0}\ \frac{e^{tan\ h}-1}{h} = \lim_{h \to 0}\ \frac{1}{h}[ 1+tan\ h(1+\frac{tan\ h}{2!}+\frac{tan^2\ h}{3!}+....\infty) - 1]\\\\=\lim_{h \to 0}\ \frac{tan\ h}{h}[ 1+\frac{tan\ h}{2!}+\frac{tan^2\ h}{3!}+....\infty]\\\\=1*\lim_{h \to 0}\ 1+\frac{tan\ h}{2!}+\frac{tan^2\ h}{3!}+....\infty\\\\=1\\

we use the following limits.

\lim_{h \to 0},\ tan\ h \to 0\ \ and\ \ \frac{tan\ h}{h} \to\ 1.
===========================

\lim_{h \to 0} \frac{e^{(x+h)^2}-e^{x^2}}{h}\\\\= \lim_{h \to 0} \frac{1}{h}*[e^{x^2+2xh+h^2}-e^{x^2}]\\\\=e^{x^2}*\lim_{h \to 0} \frac{1}{h}*[e^{2xh+h^2}-1]\\\\=e^{x^2}*\lim_{h \to 0} \frac{1}{h}*[1+2xh+h^2+\frac{(2xh+h^2)^2}{2!}+\frac{(2xh+h^2)^3}{3!}+....\infty\ \ -1]\\\\=e^{x^2}*\lim_{h \to 0} \frac{1}{h}*[2xh+h^2+\frac{(2xh+h^2)^2}{2!}+\frac{(2xh+h^2)^3}{3!}+....\infty]\\\\=e^{x^2}*\lim_{h \to 0} [2x+h+\frac{h(2x+h)^2}{2!}+\frac{h^2(2x+h)^3}{3!}+....\infty]\\\\=e^{x^2}*2x\\

the expression on the RHS simplifies as all the terms except tend to zero, as they contain h.

1 5 1
sir is there any other easier meathod its a 1 mark question.
take limit of h ->0 for derivative of numerator divided by derivative of denominator.
1) lim h->0 e^tan h * sec^2 h / 1 => e^0 * sec^2 0 = 1
Another way: given limit is the definition of e^tanh at h = 0. So evaluate its derivative and substitute h = 0.
2) same as above two methods.. given problem is same as the definition of the derivative of e^{x^2}. wrt x. so find it directly.